انا سا AN EASY Ν INTRODUCTION TO FORTIFICATION AND PRACTICAL GUNNERY. CONTAINING, 1. Decimal Arithmetic, Extraction of Roots. II. The Laws of Motion ; of Gravity. Deſcent of heavy BociesVibra- tion of Pendulums. III. Geometrical Problems, and ſome of the moſt uſeful Theorems in Geo- metry demonſtrated in a very plain and eaſy Way. IV. The meaſuring of Superficies. V. The meaſuring of Solids. Of the Strength of Beams of Timber, with Rules for cutting Scantlings for Buildings, from Mathematical Prin- ciples. VI. The Computation of Balls and Shells. VII. Logarithmical Arithmetic, VIII. PLANE TRIGONOMETRY, the Proportion of Sines, Co-fines, Tangents, &c. With many uſeful Problems for finding Heights and Diſtances, &c. copiouſly handled. IX. FORTIFICATION, explaining the Terms, with Rules for fortifying any regular Polygon, from the Square to the Decagon, and all the Sines and Angles computed by Trigono- metry, with the Operation at large. X. GUNNERY, where the Caſes are ſolved by Addition and Subtraction only, with numeral Examples ; the Rules in Words at length, whether the Projections be made on horizon- tal, or on aſcending or deſcending Planes. The Theory of Projec- tiles. Tables of Experiments of Cannons and Mortars, with ſome Obſervations. The Solution of a Problem to find the Velocity of a Bullet ſhot from any Piece of Ord- nance; with the neceſſary Tables. The Second EDITION, corrected and very much enlarged. By the Rev. F. HOLLIDAY, Miniſter of Bothamſal, and Maſter of the Free Grammar- School at Haughton Park, Nottinghamſhire. LONDON, Printed for G. ROBINSON, in Pater nofter-Row. MDCCLXXIV. 1774 HE CRITICAL REVIEW for November, 1756, gave the following account of the Firſt Edition of this work, viz." Upon the whole, we think, that the authors quoted in this work, are well choſen; the practical rules inſerted are clearly delivered, and the obſervations and examples annexed to them are pertinent and familiar. The deſign is certainly laudable, being intended, at this critical conjuncture, not only to excite in the public a deſire of at- taining fome knowledge in this neceffery ſcience, but alſo to aſlift them therein, at an eaſy expence." a White Η Williains PRE FACE. P R E F A С A GREAT part of the enſuing treatiſe was ſome years ago drawn up for my own uſe, in teaching fome young gentlemen, who had not a conveniency or opporutnity to ſearch into different authors that have treated on this ſubject : the particulars of the following ſheets are as follow; I. Decimal Arithmetic, Extraction of the Square and Cube Roots, being neceſſary to all thoſe who are inclined to pro- ceed regularly in attaining this uſeful branch of knowledge. II. Menſuration of Planes and Solids, and the method of computing the ſtrength of timber joiſts, whether oak, elm, beech, or fir; and it muſt be here ſuppoſed that the timber is homogeneous, and of the ſame goodneſs as thoſe pieces Mr. Emerſon made his experiments with, or elſe a proper al- lowance muſt be made for the defect. III. The uſual rules for finding the Weight and Diameter of Bullets, whether Lead or Iron; the requiſite charges of pow- der for any piece of artillery of any given dimenſions, and for eſtimating the quantity of powder for bomb ſhells. IV. The neceffary Problems and Theorems in the Elements of Geometry, demonſtrated in a very eaſy and ſhort way, as may be comprehended by any perſon, without the help of a maſter. V. Logarithmical Arithmetic, ſhewing the nature and uſe of Logarithms; with Plane Trigonometry folved in all caſes, il- Juſtrated with proper examples ; the uſefulneſs whereof is ſo great in all parts of the mathematics, that the moſt impor- tant and uſeful parts of knowledge would be quite loſt, if men were deſtitute of it ; for the engineer would make but a deſpicable figure when he came to take the meaſures of heights and diſtances, whether acceſlible or inacceffible, or for laying down plans of fortification on the ground, com- puting the capitals, curtains, Aanks, faces, lines of defence, &c. if he was ignorant of Trigonometry. VI. The A 2 In William tout v REF A CE. VI. The reader will find ſeveral examples for meaſuring heights and diſtances, and at one ſtation, all done by Plane Trigonometry. VII. The Theory of Praétical Gunnery, wherein the doc- trine of projectiles is illuſtrated in a multitude of the moſt uſeful problems that can occur, with numeral examples to each; and to make it more plain and uſeful to perſons of but an ordinary capacity, I have given the rules to every pro- blem in words at length, whether the projections be hori- zontal, or thoſe made on afcending or deſcending planes ; all which are deſigned more for thoſe perſons who have not ſkill and abilities in the mathematics, than for thoſe who are furniſhed with mathematical knowledge fufficient to under- ſtand the learned tracts on this ſubject, and made farther ada vances in the elements of the higher geometry. Yet, VIII. For the ſake of my more ingenious readers, I have added what the celebrated Mr. Cotes has wrote on projectiles, done into Engliſh from his Harmonia Menſurarum, where they which will find this ſubject elegantly handled by this author; to whcih I have annexed iwo general ſcholia on projectiles, of the learned mathematician Mr. Emerſon, taken from his Mecha- nics with his own conſent, and illuſtrated them with exam- ples. I took this method, becauſe many of the problems in gunnery were wrote before his Principles of Mechanics were publiſhed, and by theſe two ſcholia the folution of every pro- blem was tried, and the anſwers very exactly agreed ; ſo that I am in no concern what ſome perſons may ſay of this performance. IX. The next part treats of experiments ranged into ta- bles; theſe experiments were made with cannon and mor- tars, and calculations were made thereon from the principles of the immortal Sir Iſaac Newton, by D. Bernculli and M. Gunther : this memoir, quoted by Mr. Robins in his New Principles of Gunnerys, is printed in Latin; a particular ac- count of which is inſerted at the concluſion of this work. X. Having a little ſpare room, I have given a fluxionary folution to a difficult problem, for finding the velocity of a bullet ſhot from any piece of artillery; though this is of no material uſe in practical gunnery, yet it perhaps may pleaſe ſome of my readers, as the conclufion comes out ſo ſimple. XI. I have alſo inſerted the neceſſary tables of gun- What is ſaid of the flight of bombs in the following ſheets, is equally applicable to cannon balls; the method for directa nery, &c. ing PREF A C E. v و ing cannons, and placing them at any elevation, is by means of proper inſtruments for that purpoſe. And here I would not be underſtood to mean, that experience does always ex- actly agree with the rules given for throwing bombs to hit an object, for there often happens many unavoidable acci- dents which will cauſe ſome diference ; and yet theſe rules are not for that reaſon to be careleſly attended to, for they furnith the gunner with the readieſt and ſhorteſt methods of perfecting his practice, and will anſwer the end propoſed. Indeed a careful attention muſt be given to thoſe perſons who are daily employed in the exerciſe of artillery, and have founded their methods upon long experience; for this will always be of more weight than the moſt learned precepts. But yet, ſuch a perſon will be better able to judge by the help of theſe rules, joined to his own experience, than he can by his experience alone. As I have conformed to the method of other writers, it being the moſt eaſy and beſt adapted for beginners ; fo I ex- pect to have objections raiſed againſt me, becauſe I have made no allowance for the air's reſiſtance; now to anticipate theſe objections, What engineer computes this for Practical Gun- nery? The problem would then become very complicate, and of a very operoſe kind. The taſk is not only extrenie diffi- cult, but uſeleſs ; for to thew how far a projectile deviates from a parabolic track by the refiftance of the air, is one of the moſt intricate problems in all the doctrine of fluxions; and which is worfe ftill, it can be of no manner of uſe ; for the motion of a projectile in a reſiſting medium can be reduced to no rules, ſo as to be of any uſe in the art of gunnery. I thought, therefore, by laying down ſhort and plain rules for practice, it would be more ſerviceable to the young engineer, than to puzzle him with tedious demonftrations and perplex- ed calculations in fluxions, which would create an endleſs deal of trouble for no real advantage. It perhaps may be faid, I might have added ſome problems by fuppofing the ſhot to move in a right line to a point blank diſtance from the mouth of the piece, and then to deſcribe the curve of the parabola ; but I do not apprehend it can be of any uſe in gunnery by putting a right line and parabola together; for the curve deſcribed by a ball on this hypothe- fis, will be nothing like the curve deſcribed in the air, and therefore I have purpoſely omitted it, for the ball never dem ſcribes a right line. A 3 But vi PRE FACE. a But if any one would make a computation from the re- fiftance of the air, he will find the problem elegantly folved by fuxions at page 84, Vol. II. of my Miſcellanea Curioſa Mathematica, where it is ſaid, that the deſcription of the true curve is ſo perplex, that it is ſcarcely of any uſe; and if, ac- cording to the ſecond corollary, the reſiſtance be ſmall, as is commonly experienced by an iron ball projected through the air with a ſufficient velocity, the curve which the globe de- ſcribes in a reſiſting medium, does not differ much from a conic parabola ; yet becauſe the reſiſtance diminiſhes the ve- locity of the projection, the ordinate to the trajectory in a refifting medium will be a little leſs than the ordinate to the conic parabola ; but notwithſtanding this, the problem, by reaſon of its difficulty, is of no uſe in gunnery. The Art of Gunnery is a ſcience at all times uſeful, and much more ſo amongſt the Engliſh, who I am ſure have as great an inclination to arms and arts as any nation in Eu- rope; their matchleſs valour and bravery, their fingular pru- dence and conduct, both by ſea and land, is ſo univerſally known, that many powerful ſtates have been conſtrained to yield to them by the dint of their ſword; and as to arts and ingenious literature, they juftly claim a true title to the em- pire of human knowledge. Theſe ſciences, it is known to all, are the peculiar care of princes, that deſign to raiſe the force and power of their countries; and theſe are none of the leaſt arts, whereby the French King hath brought his ſubjects to make that figure which they at this time do; I mean, the care that he takes for educating thoſe appointed for this ſervice: he orders that ehere be profeſſors to teach theſe ſciences publicly in ſeveral parts of the kingdom, that the teachers muſt know defigning, and to teach it to their pupils, in order to lay down the ap- pearances of things in their real form and ſituation; they are to keep their ſchools open, and to read four times a week to their ſcholars, where they muſt have books and inſtru- ments neceſſary to teach their art, who have handſome fala- res from the government for that ſervice, and to teach gra- tis. The directors of hoſpitals are obliged to ſend to theſe academies every year ſeveral of their boys, to be taug and furniſhed with books and inſtruments, explained with a vaſt variety of experiments, and thereby practice and theory go on hand in hand, and receive mutual aſſiſtance from each other; and that nothing can exceed the order of theſe ſchools, the officers placed at the head of them are of the greateſt abi- 3 lity PRE FACE. vii lity and knowledge in the management of artillery, which they teach with as much method as grammar and accompts are taught in our ſchools ; and hence it is that France is well provided with ſo great a number of able and ſufficient engineers. During the time of the late unnatural rebellion, fome mathematical lectures were read in London to inſtruct officers in the art of engineering and fortification, as knowing very well the importance thereof, and this continued ſome time after the peace; and it is worthy the conſideration of the legiflature, whether the reftoring and continuing this even in times of peace, be not expedient for the bringing up en- gineers, who are ſo uſeful and valuable in the military way, and ſo difficult to be had in time of war, and ſo little danger- ous in time of peace. The arts of war and trade require much the affiftance of theſe ſciences, an officer that underſtands fortification and engineering, will, cæteris paribus, much better defend his poſt, as knowing wherein its ſtrength conſiſts, or make uſe of his advantage to his enemies ruin, than he that does not: he knows when he leads never ſo ſmall a party, what his advantage or diſadvantage in defending and attacking are, and hereby can do more ſervice than another of as much courage, who for want of ſuch knowledge, it may be, throws away himſelf, and a number of brave fellows under his command, and it is well if the miſchief reaches no further. Fortification is fo neceſſary an art, that it greatly deſerves more encouragement; for, excepting one or two academies founded on royal authority, I cannot learn that it is properly taught in any parts of this kingdom ; for moſt of thoſe gen- tlemen that make geometry their peculiar province in teach- ing it, apply it often to uſeleſs fpeculations. Now fortifica- tion is the chief art wherein geometry is uſed, and to it only is owing the great ſucceſs and improvement of military ar- chitecture; indeed, where the ground is regular, it admits but of ſmall variety; yet where the ground is made up of natural ſtrengths and weakneſſes, it affords ſome ſcope for thinking and contrivance; and what contributes much to- wards its ſtrength, with regard to its force or reſistance, are the due meaſures and proportions of its lines and angles; as it is now arrived to ſuch a degree of perfection, that in all likelihood it cannot receive any eſſential addition; ſo it is the more eaſily to be compréhended; becauſe, it ſeems im- poffible A 4 viii P R E F À CE. 66 poffible to find any thing that is not already invented for the making atttacks upon places. Earth is made uſe of for trenches, banks, and retrenchments; water, for fluices and inundations; and fire is applied various ways; below, by the contrivance of mines and their galleries, whoſe havoc is dreadful, and deftruction rapid, in a moment burſting open the bowels of the earth, rending in pieces the ſtrongeſt walls, the firmeſt towers, and blowing up whole troops of men ; above, by bombs, carcaſſes, and granadoes, directly by can- nons, mortars, &C. " Let the reader imagine to himſelf " (fays an ingenious writer) a great number of mortars play- 6 ing continually from the batteries round a befieged town, 66 and bombs like one uninterrupted tempeſt of fire falling at once on all its quarters, forcing their paflage irrefiftibly, demoliſhing buildings, and burying multitudes under the 6 rains, tearing up pavements, ſweeping whole ranks of men away, and toſſing their mangled limbs into the air, penetrating with the violence of lightning even fubterra- "neous vaults, overtbrowing, rending or fetting on fire 66 whatever they touch, their execution is experienced as fa- tal and general, as their effects are terrible and deſtruc- 66 tive.” The cunning of mankind never exerting itfelf ſo much as in their arts of deftroying one another, therefore me- thods of defence have been invented to prevent in a great mea- fure theſe effects, and this is the proper buſineſs of fortifi- cation; and unleſs mankind find out ſome other elements, there is not much probability that they can invent new me- thods, either offenſive or defenſive. It were to be wilhed that proper encouragement were al- lowed for experienced mathematicians and engineers to make all the neceſſary experiments on pieces of artillery, for this are now makes one of the moſt confiderable branches of the military ſcience, and it is abſolutely neceſſary to have it un- derſtood by many, in order to ſupport the honour of our arms, and to maintain the public ſafety of this nation; and I am ſure never more ſo than at this preſent conjuncturé, where it is ſaid from public authority, that Great Britain or Ireland is daily threatened to be invaded by one of the moſt powerful and formidable fleets and armies that France ever had; whoſe armies are in continual motion on all ſides, their coaſts towards the ocean ſwarm with troops, their roads to Flanders, Normandy, and Brittany, are continually covered with carriages laden with cannon, warlike ſtores, arms of all kinds, proviſions, &c. all the apparatus cf ſome great enter- prize, PREFACE. ptite, whoſe injuſtice as well as inſolence is circumſcribed by no bounds; who now fully faew their ambitious proje&tó, ill faith, notorious breach of treaty, and manifeſt aggravation againſt the beſt of Kings that ever ſat on the Britiſh throne. This attempt of invading theſe kingdoms in revenge for his Majeſty's generous and fteady conduct in maintaining the rights and poffeffions of his crown and ſubjects, againſt the unprovoked aggreſſions and hoſtilities firſt begun on the part of France, is ſo unjuſt, daring and infolent, as not to be pa- ralleled in any hiſtory. But- To conclude, as the practice of Sea Gunnery is extreme eaſy, their marks being moſtly point blank or very near it, and that aſcents and deſcents bear no conſiderable fare here, I have taken no notice of it, and have omitted the de- monſtrations of the rules in gunnery, in order to render the whole as conciſe, and deliver it with as much plainneſs as my own abilities and the nature of the ſubject would permit me; and I hope the candid and impartial reader will excuſe what is amiſs, and amend thoſe errors which unavoidably attend things of this nature, or may happen by reaſon of the author's great diſtance from the preſs; for he has no other defign in publiſhing this book, than to be ſerviceable to his own countrymen, for whoſe benefit he heartily wiſhes it was much more complete than it is. Sept. 4, F. HOLLIDAY. 1756. P. S. From the favourable reception the former edition of this work has met with, and the opinion which ſome good judges of the ſubject have been pleaſed to form of it, the au- thor thinks himſelf authorized to conſider his plan as right. This, joined with his deſire to oblige thoſe who wiſh to ſee the ſubject more complete, induced him with the utmoit care to reviſe and improve it. In this ſecond edition are made many alterations and additions in ſeveral parts of the book, as the laws of motion, the laws of gravity, deſcent of bodies, the vibration of pendulums. To the geometry are prefixed the definitions, which were omitted in the former edition. At the deſire of ſome friends, I have enlarged the number of propofitions in the menſuration of planes and ſolids; and have treated largely of the ſtrength of beams of timber, with rules for cutting the proper dimenſions of ſcantlings, founded on mathe- X PRE FACE. mathematical principles, delivered in ſuch a manner as to explain them to the meaneſt capacity. To the trigonome- trical part, I have given the proportion of fines, co-fines, tan- gents, &c. and ſeveral other uſeful and curious propofitions, too long to be mentioned here. And farther, to render the book as uſeful as I could, there is now added a ſhort intro- duction to fortification, the definitions are according to the beft French writers, whom I conſulted ; alſo I have given the operation at large, for computing all the lines and an- gles of regular fortified polygons, as the ſquare, pentagon, hexagon, and octagon, from trigonometry, as laid down in this treatiſe. In gunnery, or the doctrine of projectiles, the reader will meet here and there inſerted farther improve- ments, and more examples for illuſtrating the learned Mr. Emerſon's two general ſcholia, on the motion of projectiles, with a table of theorems for horizontal projections; and it may here be obſerved, that Propofitions I. II. VII. and XI. are the moſt uſeful for practice, which are few and very eaſy. Theſe, it is prefumed, will be acceptable to gentlemen in the army, which are peculiarly adapted to afiſt thoſe who deſire to be acquainted with ſo uſeful a ſcience. Though men in private life may be allowed to neglect theſe ſciences, yet perſons of eminent ſtations, who have the conduct of ſo- ciecies and ſtates, may by no means be admitted to do it; for not only military gentlemen, but all others who would ſpeak properly on this ſubject, and underſtand what they read in the public news-papers, ſhould be acquainted with this ne- ceſſary branch of ſcience. If that which is here delivered prove uſeful, it will be a pleaſure to him, who, upon all occaſions, is glad of pro- moting real and uſeful knowledge. May 4, 1772. F.H. Τ Η Σ THE C Ο Ν Τ Ε Ν Τ T N T S. panel 5 6 ibid. 7. 9 12 ibid. 13 17 21 24 27 29 30 of decimal arithmetic Addition Subtraction Multiplication Diviſion The rule of three in decimals Compound rule of three direct Compound rule af three inverſe Extraction of the ſquare root To extract the cube root The laws of motion Of gravity, defcent of heavy bodies Vibration of pendulums Geometrical definitions Geometrical problems Some of the moſt uſeful theorems in geometry demonſtrated To multiply feet and inches To meaſure a ſquare To meaſure a parallelogram Of gunter's chain To find the area of a triangle To find the area of a trapezium The properties of a circlc, &c. &c. To determine the ſize of water pipes To find the area nf an ellipſis To find the periphery of an ellipſis An elliptical problem To find the area of a ſemi-ellipſis To find the area of a quadrant of an ellipſis To find the area of regular polygons 36 52 55 56 ibid. 59 60 62 65 69 ibid. 70 72 ibid. 73 TO 85 96 [xii Page To find the ſurface of arches of bridges 75 To nieaftere irregular figures ibid. To take offsets of crooked rivers, hedges, &c. and caſt up their contents 77 To find the folidity of a cube 80 To find the ſolidity of a parallelopipedon ibid. To find the Solidity of a triangular priſm 81 To find the ſolidity of a cylinder 82 To find the ſolidity of a pyramid ibid. To find the folidity of a body bounded by four triangular faces 83 To find the ſolidity of a fruftum of a pyramid 84. To find the folidity of a cone To find the Solidity of a fruſtum of a cone ibid. To find the ſolidity of a globe or ſphere 86 To find the ſolidity of Noping bodies 87 To find the ſolidity of a rampart 91 To find the quantity of earth dug in making a ditch with ſloping fides 95 Of the frength of beams of timber Rules for cutting ſcantlings, or finding the proper dimenfions of tie beams, bridging joiſts, trimming joills; &c. The computation of balls and ſhells 109 A table Shewing the queight of any ball of iron or lead from two to leven inches diameter 115 A table of gunnery 116 Experiments on gunpowder from the French of M. D'Arcy, and M.Le Roy 117 To find the cubic inches in the hollow part of any bomb, and the quantity of powder io fill it 119 A table of quantities of powder for filling mells, with weight of compoſition requiſite io fill each fuſe, length of the fuſe, and number of ſeconds each fuſe burnt A table of mortars and the dimenſions of shells A table ſhewing the requiſite weight of powder for mortars To find the number of cannon balls in any pile, whether ſquare, oblong, or triangular, finiſhed or unfiniſhed 123 LOGARITHMICĂL ARITHMETIC To find the logarithm of any whole number ibid. To find the logarithm of any mixed number 129 To find the logartthm of any number, that exceeds the limits of the tables ibid. Any logarithm being given to find the number anſwering thereto 130 Multipli- 98 I20 121 122 128 I xiii a Page Multiplication by logarithms 131 Diviſion by logarithms 132 To ſquare, cube, &c. any given number by logarithms 134 To extract the roots in logarithms ibid. To work the rule of three logarithmically ibid. PLANE TRIGONOMETRY. Definitions 135 The proportion of fines, co-fines, tangents, &c. 136 To find the natural fines, c9-fines, tangents, &c. ibid. To find the logarithmic fines, co-fines, tangenis, &c. 137 To find the fine, co-fine, tangent, &c. to 90 degrees of the in- termediate parts of a minute ibid. To find the arc, from the fine, co-fine, &c. 139 Given the radius and arch of a circle, in degrees and minutes, to find the length of that arch 140 Given the ſum of two angles, and the product of their fines, to find the angles ſeparately 141 Axioms of trigonomeiry 142 The ſolution of the ſeveral caſes of right angled triangles 144 The ſolution of oblique triangles 153 To meaſure an acceſſible height 160 To find the breadth of a river 161 To meaſure an inacceſſible height 162 The meaſuring of diſtances at one flation 163 To take a long diſtance on a level plane, without the help of any inftrument, unleſs a faff to let off a right angle 172 The meaſuring the diſtances at two ſtations ibid. Two curious problems 173 To find the diſtance of places by firing of cannon 179 Aiable of the angles which every point of the compaſs makes with the meridian 181 Problems fbewing its uſe 182 FORTIFICATION. Definitions 186 Explanation of ihe terms of a fortified place 191 To fortify a ſquare 192 To fortify a regular pentagon 196 To fortify a regular hexagon 206 To fortify a regular očlagon A table of the lines and angles of a regular fortified polygon from the ſquare to the decagon 215 A table of dimenſions for fortifying the body of the place 216 To draw the profile of a regular fortification ibid. To repreſent the profile in perſpective 218 eſſay on encampments 219 GUNNERY. Definitions Of 1 a 211 221 [ xiv ] Page Of the parts of cannon 230 Projections on the plane of the horizon 231 Projections on aſcending planes 240 Projections on deſcending planes 255 The theory of projectiles 262 Explanation of Mr. Emerſon's two ſcholia on projectiles 266 A table of theorems for horizontal projections 270 Two tables computed from Sir Iſaac Newton's principles, and applied to the motion of projectiles by D. Bernoulli A problem with its folution, in order to determine the velocity of a bullet ſhot from a piece of ordnance 279 276 An explanation of the SYMBOLS uſed in this treatiſe. + fignifies addition, as A+B denotes B added to A. fignifies ſubtraction, as A-B denotes B to be taken from A. x ſignifies multiplication, as A x B denotes A multiplied by B. fignifies equality, as A=A or A is equal to A. ✓ fignifies the ſquare root, as ~ A the ſquare root of A, or ✓144 the ſquare root of 144 or 12. Log. fignifies the logarithm of. : , : : fignifies proportion, or rule of three, thus 2:4::6: 12, that is, as 2 is to 4. fo is 6 to 12. A fignifies triangle, as A ABC denotes the triangle ABC. fignifies angle, as 2 A fignifies the angle A. $. fignifies fine, as s. ABC denotes the fine of the angle B. $. fignifies tangent, as t. ABC denotes the tangent of an- gle B. songs fignifies the ſquare, as CD denotes CD ſquared. fignifies parallelogram, as DB the parallelogram DB. º fignifies dgrees, as 24° fignifies 24 degrees. fignifies minutes, as 5 denotes 35 minutes. 1 C Glit-Frillian Р ART 1. OF DECIMAL ARITHMETIC. DEFINITIONS. A FRACTION is called vulgar or decimal, according as it is divided, as is called a vulgar fraction, the up- permoſt number is called the numerator, and the under- moſt the denominator. A decimal fraction ſuppoſes one of any thing whatever to be divided into 10 parts, 100 parts, 1000 parts, &c. and hath always a dot fet before it as .5 is called 5 tenths, and in vulgar fractions, but as decimals are the eaſieſt, I ſhall ſay nothing of vulgar fractions here. As places in whole numbers increaſe by tens, ſo do places in decimals decreaſe by tens; for as the firſt, ſecond, third, &c. place above that of units is tens, hundreds, thouſands, &c. ſo the firſt, fecond, third, &c. place below that of units is tenths, hundredths, thouſandths, decreaſing in a tenfold propor- tion, as is manifeſt from the following table. To a 6 Hundreds of thouſands 7 Millions. 5 Tens of thouſands. 4 Thouſands. 3 Hundreds. 3 Hundredth parts. 2 Tenth parts. 2 Tens, 4 Thouſandths. Ten thouſandths, 1 Unit, 5 "syspuejnoq; paipunh 9 7 Millionths. B So 2 DECIMAL FRACTIONS. P.I. so { 1.175 25} fignifies -52 5 Tenths. . 25 Hundredths. 175 Thouſandths. Cyphers fet after decimals do not alter their value, as .8 and 8000 are the fame, viz. only 8 tenths : but when cyphers are ſet before decimals, they remove them farther from the decimal point, and therefore decreaſe them, as .08 is leſs than .8, fo .008 is leſs than ..8, conſequently, each removal of a figure into a place forward makes it ten times leſs than it was before. To reduce Vulgar Fractions into Decimals. RULE. Add as many cyphers to the numerator as you intend deci- mals, and divide by the denominator, the quotient is the de- cimal required. Examples. 1. Reduce of any thing into a decimal. 4) 1.00 (-25 the decimal. 8 20 20 2. Reduce of any thing into a decimal. 2) 1.0(.5 the decimal. TO a 3. Reduce of any thing into a decimal. 4) 3.00 ( -75, the decimal. 28 20 20 So the quarter, half, and three quarters of any thing are .25; 5; and .75 decimally. 4. Reduce P.I. DECIMAL FRACTIONS 4 Reduce into a decimal. 1 2 2 7) 2.00000(.28571 the decimal of $. 14 60 56 40 35 elembe 50 49 quan 20 ) m IO 7 3 32 5. Reduce into a decimal, 5 32 ) 1.00000(.03125 96 40 32 80 64 160 160 Note. Here I annexed five cyphers to i the given numerator, as I intended to have five places in decimals, and there ariſe but four in the quotient, I muſt ſupply ſuch defect by prefix- ing as many cyphers on the left-hand of the firſt figure in the quotient as there want places ; as in the example above there is prefixed a cypher to make the quotient compleat, becauſe I annexed five cyphers to the numerator. 6. Reduce 6s. 8d. into the decimal of a pound. Thus, 6 s. 8d. is = 80 d. and a pound = 240 d, Then 240) 80.00000 (33333 is the decimal. В 2 7. What 4 P.I. DECIMAL FRACTIONS, 7. What is the decimal of 5 cwt. I qr, ilb. the integer being a tun weight ? Thus, a tun weight = 2240 lb. and 5 cwt. I qr. 11lb.= 599 lb. Then 2240) 599.0000 (.2674, the decimal. To find the Value of any Decimal Fraction. RULE. Multiply your given decimal by the parts of the next infe- rior denomination that is equal to the integer, the decimal dotted off gives the part thereof; and if there be any remain- der, multiply it by the next inferior denomination, &c. till all is done. Examples. 1. What is the value of.875 of a pound Sterling ? Thus £ .875 20 S. 17.500 12 d. 6.000 Anſwer 17 s. 60. 2. What is the value £o:9874? Thus £.9874 20 19.7480 12 8.9760 4 Anfwer 19 s. 8.904d. 39040 3. What is the value of.5875 of a tun weight? Thus •5875 Tun. 20 II.7500 4 of Anſwer I cwt. 39rs. 3.0000 4. What P.I. DECIMAL FRACTIONS. 5 4. What is the value of .3712 of a hundred weight? Thus 4 •3712 cwt. 1.4848 28 38784 9696 13.5744 16 34464 5744 9.1904 Anſwer. I qr. 13 lb. 9.1904 oz. ADDITION of DECIMALS. RU LE. LE ET the numbers be placed according to their value, as units under units, tens under tens, &c, in whole num- bers; and in decimals tenths under tenths, hundredths under hundredths, &c. then work as in whole numbers, always re- w membering to make as many places of decimals in the fum total, as the greateſt number thereof were in your example. Examples. £ Yards. £818.4 6.97 .372 21.648 3.81 .281 6.5 •54 .846 19.8168 .61 75 1.82 .6 16.614 3.25 •583 48.94 72 1108 £3.432 185.6296 835.40 Or £3:8:73.72 Or 185 yds. 2 qrs. 2 na. &c. Or £835: 8 B3 SUB 6 P. I. DECIMAL FRACTIONS SUBTRACTION. Place your numbers as directed in addition, then work ag in whole numbers. Examples. From •591 .6198 I. 54 Take .397 .1499 0.987 6.9843 Remains 194 .4699 0.013 47.0157 MULTIPLICATION. Work in multiplication as in whole numbers, always re- membering to ſet off as many decimals in the product as there are decimals both in your multiplicand and multiplier, and if you have not a ſufficient number in your product, then prefix as many cyphers as you want. Examples. 4918 .37 37 .34.16 23912 10248 34426 14754 .126392 18.1966 37.241 12,23 .131461 .2132 111723 74482 262922 394383 131461 262922 74482 37241 455.45743 .0280274852 DIVE P. I. 7 DECIMAL FRACTIONS. DIVISIO N. Divide as if all were whole numbers, annexing cyphers to the dividend, if need be; and when you have finiſhed the operation, point off for decimals in the quotient, ſo many places as the decimal parts uſed in the dividend exceed thoſe of the diviſor, and thoſe to the left, if any, are whole num. bers. If the places in the quotient are not as many as the above rule requires, then ſupply the defect with cyphers on the left hand of the quotient. Examples in all the Varieties. 246) 160.884 (.654 quotient. 1476 13 28 12 30 984 984 246) 16 08846.0654 1476 1 328 1 230 984 984 2.46) 1608.84 (654 1476 1328 1230 984 984 B4 2.46 DECIMAL FRACTIONS, P. I. 2-46) 16.0884 ( 6.54 1476 I 328 1 230 984 984 2.46 ) 160884.00 (65400 1476 1328 1230 984 984 oo .0246) 160.8840( 6540 1476 1328 12 30 telo 984 984 0246 ) .160884 (6.54 1476 1328 1230 984 984 Tbe P.I. 9 DECIMAL FRACTIONS. The Rule of Three in Decimals. State your queſtion as in whole numbers, and work as aforeſaid. Examples. If a pound and half of butter coſt a groat, what will two pounds coſt? Thus 1b. d. Ib: If 1.5 4 4 d. 1.5) 8.000 (5:33 75 4 2 50 1.32 45 50 45 Anſwer 54.32 d. 5 Two men bartered ; A had 40.7 yards of fhalloon, for which В gave him 25.6 ells of Holland, at 4.6. 6 d. per ell, what is the ſhalloon a yard? Ell. Ells. If I 4.5 4.5 s. 256 128 1024 s. 115:20 If IO P. Z DECIMAL FRACTIONS. Yards. S. If 40.7 Yard. - I 115.2 I S. 40.7) 115.2000 2.83 814. 12 3380 d. 9.96 3256 4 I 240 I 221 3.84 Anſwer 2 s. 91.8 d. per yard. 19 A owes B £296 : 17s. but not being able to pay the whole, he compounds for 7 s. 6 d. in the pound, what muſt B re- ceive for his debt? Firſt, the decimal of 17 s. is £.85 s. £ £ If I 7.5 296.85 7.5 S. 148425 207795 $. 2226.375 12 4.500 4 Anſwer 2226 s. 41.6d or £111: 6:41 2.000 Note, When more requires leſs, that is, when the third term is greater than the firſt, and requires the fourth term to be leſs than the ſecond, or when leſs requires more, that is, when the third term is leſs than the firſt, and requires the fourth term to be greater than the ſecond, then the propor- tion is inverſe; and to find the fourth term, Multiply the firſt and ſecond terms together, and divide that product by the third term, the quotient is the fourth term, or anſwer to the queſtion. Examples 3 P. I. DECIMAL FRACTIONS. Examples. There is proviſion enough in a caſtle to ſerve a garriſon of 200 ſoldiers for a year, and it was reported that an enemy were coming to attack it, the garriſon was augmented to 500 men more, how many days will the proviſion ſerve theſe 500 men Men. If 200 Days. 365 200 Men, 500 slco) 7301000 146 days, anſwer, A certain garriſon, containing 4000 men, hath proviſion fufficient for ſix weeks and four days, but being about to be beſieged by the enemy, and no aſiſtance to be got to relieve it in leſs than two months; how many men muſt the gover- nor diſcharge out of the ſaid garriſon, that the ſaid proviſion may ſerve the remainder four months two weeks? Firſt, the decimal of four days, a week the integer, is .57 For 7) 4.00 (:57 Weeks. If 6.57 Men. - 4000 Weeks, 16 4000 16) 26280 ( 642 will ſerve, 16 102 96 68 64 From 4000 Take 1642 40 32 2358 men to be dir- charged. 8 Com- 12 P.I. DECIMAL FRACTIONS. Compound Rule of Three Direet. In the compound rule of three direct there are five nmbers given to find a fixth, which is to the third as the product of the two laſt to the product of the two firſt. Multiply the firſt and ſecond together for a divifor, and the fourth, fifth, and third terms together for a dividend the quotient ariſing by dividing the dividend by the divi- for is the anſwer. Example. If 12 men in 8 days dig 48 cubic fathoms of earth to make an entrenchment, how many fathoms at the ſame rate will 28 men dig in 4 days ? Men. Days. Fath. Men. Days. 8 28 4 8 4 If 12 48 96 divifor II2 48 896 448 96) 5376 (56 fathoms. Anf, 480 576 576 Compound Inverſe Rule of Three, Is to find a fixth number from five given numbers, which is to the third as the product of the two firſt is to the product of the two laſt. To find the fixth, multiply the product of the two firſt by the third, and this product divide by the product of the fourth and fifth, the quotient is the fixth, or anſwer. Example P. I. 13 DECIMAL FRACTIONS. Example In a garriſon containing 3000 men there was, at the time of their being beſieged, a proviſion of bread in the magazine fufficient to ſerve the full complement, viz. 4000 men, juſt three months, at 20 ounces daily allowance per man: but the governor being determined to defend the garriſon fix months, deſires to know how much bread each man may be allowed a day for that purpoſe? Men. Ounces. If 4000 Months. - 3 Months, 6 - 20 Men. 3000 6 - 3 18000 12000 20 18100) 240100o ( 131 ounces for each man a day. 18 18101 60 54 6 I. 2. 3. 4. 5. 6. Extraction of the Square Root. NY number multiplied into itſelf is ſaid to be ſquared, А or the ſquare of that number; thus 6 multiplied into itſelf, or 6 x6 = 36, the ſquare of 6; whence the following table : Roots 7. 8. 9. Squares 1. 4. 9. 16. 25. 36. 49. 64. 81. RU L E. When any number is given to be extracted, point it off in every other place, beginning at unity ; then find the firſt figure in the root, whoſe ſquare ſhall be equal to, or neareſt leſs than the figure or figures to the firſt point, and then ſub- tract that ſquare from the firſt point, and to the remainder bring down the next point, and call that the reſolvend. Then double the quotient or root, and ſet it for a diviſor on the left- 2 14 The SQUARE ROOT. P.I. Jeſt-hand of the reſolvend, and ſeek how oft the diviſor is con- tained in the refolvend, referving always the units place, and put the anſwer in the quotient or root, and alſo in the units place of the diviſor; then multiply the divifor by the figure laſt put in the quotient or root, and ſubtract the product from the reſolvend, as in common diviſion, and bring down the next point to the remainder, and this is the reſolvend; and proceed as before till the operation is finiſhed. Examples. What is the ſquare root of 116964? Firſt let it be pointed thus, 116964; then find the feareft {quare number to u the firſt period, as 9, &c. Thus: 116964 ( 342, root. 9 64.) 269 reſolvend. 648 256 1368 1026 682) 1364 reſolvend 1364 116964 proof. For 342 by 342 What is the ſquare root of 13169641? Thus, 13169641 ( 3629 root. 9 66) 416 396 722) 2096 1444 7249 ) 65241 65241 O When any whole number is propoſed to extract the ſquare root, and there be a remainder, the root may be found to any degree of exactneſs, by adding two cyphers to the re- mainder, and proceeding as before. What P.I. The SQUARE ROOT. I5 What is the ſquare root of 37468 ? 37468 ( 193-566, &c. I 29 ) 274 261 38 3) 168 149 3865). 21900 19325 38706) 257500 232236 387126) 2526400 2322756 203644, &c. When a mixt number or decimal is to be extracted, point every other figure as before, beginning at the units place in whole numbers, but in the decimal always begin in the ſe- cond or hundreds place, and if you have not a full period then add a cypher to compleat it, and when you have not as many figures in the root as dots or periods, then prefix a cypher or cyphers to anſwer them. What is the ſquare root of 523.176? 523.1760) 22.87, &c. 4 42 ) 123 84 448 ) 3917 3584 4567) 33360 31969 1391, &c. What 16 The ŚQUARE ROOT. P.I. What is the ſquare root of .143219? .143219(-3784. &c. 9 67) 532 469 748) 6319 5984 7564) 33500 30256 3244, &c. What is the ſquare foot of 7 7.000000 ( 2.645, &c. 4 - 46)300 276 524) 2400 2096 5285) 30400 26425 3975, &c. What is the ſquare root of .0007612816? 0007612810(-02759 4 47)361 329 545) 3228 2725 5509)50316 49581 735 P.I. 17 The CUBE ROOT. To find the ſquare root of a vulgar fraction, reduce it into a decimal, and extract the ſquare root of that decimal, and the thing is done. What is the ſquare root of 4 or ? Thus for in decimals is .5 then.5000 ( 707, &c 49 1407) 10000 9849 151, & C. To extract the Cube Root. RU L E. Firſt point off from the unity place every third figure, find the firſt figure in the root by the following table, fubtract it from the firſt period, augment the remainder by the firſt figure in the next point, this call the dividend; divide this dividend by the triple ſquare of the whole root laſt found, the quotient figure found annex to the root; cube the root thus found, and fubtract it from as many points as you have brought down; to the remainder bring down the firſt figure of the next point for a new dividend, divide this new divi- dend by the triple ſquare of the root, the quotient figure annex to the root again, cube the whole root, and ſubtract it from as many points as you have brought down, and proceed in this manner till the work is ended. The TABLE of POWERS. Roots 8 I 2. 3 4 5 6 7 9 Square 1 4 9 16 25 36 49 64 81 Cubes I 8 27 | 64 64 125 216. 343 512 729 C Thus 18 Ρ.Ι. The CUBE ROOT, Thus to extract the cube root of 13312053, the number is firſt to be pointed after this manner, viz. 13312053, then you are to write in the root the figure 2, whoſe cube 8 is the next leſs cube to the period 13, which is not a perfect cube number, and having fubtracted that cube according to the rule, there will remain 5, which being augmented by the next figure of the reſolvend 3, and divided by the triple ſquare of the root 2 or 12, by ſeeking how oft 1 2 is contain- ed in 53, it gives 4. for the ſecond figure of the root, but fince the cube of 24, viz. 13824 would come out too great to be ſubtracted from 13312 that precedes the ſecond point, there muſt only 3 be wrote in the root, then 23 cubed is 12167, which taken from 13312 will leave 1145, which augmented by the next figure of the reſolvend o, and divided by the tri- ple ſquare of the quotient 23, viz. 1587, it gives 7 for the . third figure of the root; then 237 cubed and ſubtracted from: the three periods or the whole reſolvend, there remains o. Whence the true root is 237. See Sir Iſaac Newton's Arithe metica Univerſalis. Examples. 13312053 (237 Rool 2 X 24X3=12)53(4 or 3. Subtra& cube 1 2167. 23 X 2=529 X3=1587)1145007 Subtract cube 13312053 Remains O What is the cube root of 27054036008? 27054036008 ( 3002 Root. 27 3X3=9X3=27)054262 3002 cubed is 27 054036008 o What P.I. 19 The CUBE ROOT. What is the cube root of 122615327232 ? 122615327232 ( 4968 Root. 64 4x=16x3=48) 586 ( 9 49 cubed, fubtract 117641 49x49=2401+3=7203) 49743 (6 Subtract 496 cubed, viz. 122023936 496 x 496=246016x3=738048) 5913912 (8 Subtract the cube of 4568, viz. 122615327232 Remains O When the reſolvend has à remainder, you may carry on the root to any degree of exactneſs by adding three cyphers for each new period you intend to carry the root to in places of decimals; and always remember to point off every third figure from the tenths place in decimals, the whole numbers as before. What is the root of 61218.00121? 61218.001212 ( 39.41 root 27 3x3=9x3=27) 342 ( 9 39 cubed 59311 39 x 39=1521 X3=4563) 19070 ( 4 394 cubed 61162984 394/qd =155236 x 3=465648 ) 550172 ( 1 3941 cubed, is 61209566621 Remainder 8434599 C 2 What 20 P.I. The CUBE ROOT. What is the cube root of .0069761218? .006976121800000 ( .19107 J. IXI=IX3=3) 59 (9 19 cubed 6859 19x19=361 X3=1083) 1171(I 191 cubed 6967871 191 X 191=36481 X3=109443 ) 82508 (0 1910 cubed 6967871000 1910ſqd. =3648100 X3=10944300 ) 82508000 ( 7 19107 cubed, is 6975534818043 Remiander 586981957 PART II. [ 21 ] PART II. Τ Η Ε LAWS OF M O T I O N. T . DEFINITION S. 1. THE place of a body is that part of ſpace, which a body occupies. 2. Motion is a continual change of place. In motion there are three things to be conſidered, the body which is moved, the ſpace which is paſſed over or deſcribed, and the time in which it is deſcribed. 3. The direction of motion is a right line which a body in motion deſcribes, or endeavours to deſcribe. 4. Eguable or uniform motion is that, by which a body in motion deſcribes or paſſes over equal ſpaces in equal times. 5, Accelerated motion is that, by which a body in motion deſcribes continually greater ſpaces in equal times; or which is continually increaſed. 6. Retarded motion is that, which continually decreaſes in equal times; and if the increaſe or decreaſe of motion be equal in equal times, the motion is then ſaid to be equally accelerated, or retarded. 7: Velocity or celerity is that affection of motion, whereby a a body in motion paiſes over, or deſcribes a given ſpace in a given time. Ler V expreſs the velocity, T the time, and S the face; when it is faid, the velocity is dire&tly as the ſpace, and in- verſely as the time; or the velocity is as the ſpace directly, and time reciprocally, it is expreſſed thus V :: fo also the , : time T : S, that is, the time is as the ſpace dire&ly and velocity inverſely. Hence theorems may be determined for . the time, velocity, ſpace pafied over, in equable and acce- lerated motion, as follows. C3 Let 22 The LAWS of MOTION. P.II. Let Q and q expreſs the quantities of matter in two bodies. V and v, their reſpective velocities. M and m, their momenta, or quantity of motion S and S, the ſpaces deſcribed or paſſed over in the times T and t reſpectively. Then 1. If the velocities are equal, the momenta will be as the quan- tity of matter in the moving bodies, that is If V =0 Then M : m :: Q:9 Whence Mq = me M = And I. - 2. 9 addi-qizajdete 3. m 4. Q = II. If the quantities of matter in the moving bodies are equal, the momenta will be as their reſpective velocities. That is If Q = 9 Then M : m :: V v. III. The ratio of the momenta is in complicate ratio of the quantities of the matter and the velocities. That is M : m ::Q X V:9 X V. IV. The ratio of the velocities is compoſed of the direct ratio of the momenta, and the inverſe ratio of the quantities of matter. That is V:0:: 9 x M:Q x m. V. If the momenta are equal, the velocities will be inverſely as the quantities of matter in the moving bodies. That is If M = m Then V:0::2: Q. VI. Ρ. ΙΙ. 23 The LAWS OF MOTION. VI. If the times are equal wherein bodies move, the ſpaces deſcribed will be as the velocities with which they move That is IF T = 1 Then S:s :: V : v. VII. If the velocities are equal, the ſpaces deſcribed by the moving bodies, will be directly as the times. That is If V V Then S:si T:t VIII. In all caſes, the ratio of the ſpaces deſcribed is in the com- plicare satio of the times and velocities conjointly. That is S:s :: T * V:t x v. IX. If the ſpaces deſcribed are equal, the velocities are in the reciprocal ratio of the times. That is If S=S Then V:v ::t: T. X. The ratio of the times is compounded of the direct ratio of the ſpaces deſcribed, and the reciprocal ratio of the veloci- ties. That is T::: S XV:s X V. XI, The ratio of the velocities is compounded of the direct ratio of the ſpaces deſcribed, and the reciprocal ratio of the times. That is V: 0 :: 1 x S:T S. . XII. The ratio of the times is compounded of the direct ratio of the ſpaces deſcribed, divided by the velocitiese T:+::$ : t СА There S o s 24 P. II. The LAWS of MOTION. Theſe are the laws of motion deduced from thoſe general laws of Sir Iſaac Newton, in his Principia, and are elegantly demonſtrated in Mr Emerſon's Mechanics; whence equations may be deduced, as given in the firſt. SCHOLIU M. Motion is meaſured by the velocity and quantity of matter. Therefore if a ball of iron, and a ball of wood of the fame bigneſs be projected with the ſame velocity, there will be more motion in the ball of iron, than in that of wood ; ſo likewiſe, if two equal leaden balls, the one ſolid, the other hollow and empty, be moved with the ſaine velocity; the ſolid ball will have more motion than the hollow one, and will ſtrike a body againſt which it is thrown with greater force; and the quan- tity of matter, which is properly contained in any body, is to be determined by its weight: wherefore the quantity of mo- tion is not to be meaſured by the velocity and bigneſs, but by the velocity and weight of the body in motion, which is carefully to be obſerved ; and all bodies refift in proportion to their denſity, that is, to the quantity of matter contained in them. Whence we have the following concluſions. ſpace 1. Velocity = 2. Space = velocity multiplied by the time. ſpace ſpace of deſcribing any ſpace. velocity 1612 Momentum, or the quantity of motion, is compounded of the velocity and quantity of matter, or weight. Whence 4. Momentum = velocity multiplied by the weight. 5. Weight velocity 6. Velocity = weight time 3. Time momentum - momentum Of Gravity, Defcent of beavy Bodies. The velocities of deſcending heavy bodies, are proportional to the times from the beginning of their fall. This follows, becauſe the action of gravity being continual, in every ſpace of time, the following body receives a new impulſe, equal to what it had before in the ſame fpace of time received from the firft P. II. DESCENT of Heavy BODIES. 25 a firſt power, viz. In the firſt ſecond of time, a body hath acquired a velocity of 32 , for an heavy body let fall from any height near the ſurface of our earth, deſcends in a ſecond of time 16 ti feet, and were there no new force, it would continue to deſcend at that rate with an equable motion : but in the next ſecond of time, the ſame power of gravity conti- nually acting thereon, fuperadds a new velocity equal to the former, ſo that at the end of two ſeconds, the velocity is double to what it was at the end of the firſt, and after the fame manner, may it be proved to the triple at the end of the third ſecond, and foon; wherefore the velocities of falling bodies are proportional to the times of their falls. Cor. 1. The ſpaces deſcribed by the fall of a body are as the ſquares of the times from the beginning. Cor. 2. The velocities acquired by bodies falling are in proportion to the ſquares of the times in which they fall; therefore the velocity is as the ſquare root of the height fallen. Cor. 3. All bodies deſcending or ałcending gain or loſe equal velocities in equal times, and whatever a velocity a falling body gains in any time; if it be projected directly upwards, it will loſe as much in an equal time; therefore if a body be projected upwards with the velocity it acquired by falling in any time, it will in the ſame time loſe all its motion; hence alſo bodies projected upwards loſe equal velocities in equal times. Cor. 4. If a body be projected upwards with the velocity it acquired in falling, it will in the ſame time aſcend to the height it fell from ; and deſcribe equal ſpaces in equal times, both in aſcending and deſcending, but in an inverſe order; and will have the ſame velocity at every point of the line deſcribed; for gravity acts inceſſantly and at every inftant of time from the very beginning to the end of the fall. Cor. 5. If bodies be projected upwards with any velocities, the heights of their aſcents will be as the ſquares of the times of their aſcending. All theſe things would be true, if it was not for the reſiſtance of the air, which is very great in ſwift motions, and has a great effect in deſtroying the motions of bodies. Cor. 6. The forces requiſite to throw equal bodies to dif- ferent heights, are to each other, as the ſquare roots of heights. Therefore any diſtance or depth may be found by this. PRO- 26 P. II. DESCENT of Heavy BODIES. feet - PROPOSITION. I. Given the time in ſeconds of a body deſcending to find the ſpace it has fallen through. RUL E. Multiply the ſquare of the ſeconds, the body took in falling by 16 6 feet, and the product gives the ſpace it has deſcribed, Example Suppoſe a bullet was 4 ſeconds in falling from the top of a ſteeple, how high is it? " fq. time 4 X 4 16 X 161 = 257 feet. Thence may be determined the height of any ball that is projected perpendicular to the horizon. For half the time of the ball's continuance in the air, is the time of its deſcent. Example. Suppoſe a cannon placed perpendicular to the horizon, and then fi red off in that direction, it was found by obſervation, that the ball ſtaid in the air 23 ſeconds, how high did the ball aſcend? 2)23 ſeconds 11.5 X 11.5 = 132.25 + 16 = 2127.02 feet the height the ball afcended: but becauſe of the reſiſtance of the air it will retard the motion. PROPOSITION IT. Given the height of any place, to determine in what time a heavy body let fall from the top will reach the bottom. RUL E. Divide the given height in feet by 16tz or 16.09 and the quotient is the ſquare of the time, whole ſquare root is the number of ſeconds required. Example Admit there is a precipice 2127.02 feet high, in what time would a bullet be falling to the bottom ? 16.09 ) 2127.02 ( i 32.19, 11.5 ſeconds. It P. II. 27 Of PENDULUM S. It is eaſy to demonſtrate, that the time of the leaſt vibration of a pendulum is to the time of the fall of a body from the height of half the length of the pendulum, as the circum- ference of a circle, toits diameter very near. Whence it follows, As the ſquare of the diameter, is to the ſquare of the cir- cumference; ſo is half the length of a pendulum vibrating feconds, to the ſpace deſcribed by the fall of a body in the firſt ſecond of time. The length of a pendulum vibrating ſeconds is found to be 39.13 inches, and the deſcent in that ſecond will be found to be 16+feet. Now 7.17 = time of falling through 2x39.13. Whence by Cor. 1. we have 14to? 2 X 39.13 :: 1 : 39,13 * 3.1416 X = 193.096 inches == 16.083 feet. 4 2 3 2 OF PENDULUM S. The lengths of pendulums are always accounted from the point of ſuſpenſion to the center of ball or bob, but if a large ball, then from the point of ſuſpenſion to the center of oſcil- lation ; and their lengths are to each other as the ſquares of the times of vibration performed in one and the ſame time. PROPOSITION I. PROB. To find the length of a pendulum which ſhall make any aſſigned number of vibrations in a minute. RULE. Say, as the ſquare of the aſſigned number of vibrations, is to the ſquare of 60, the ſeconds in a minute; ſo is 39.13, the the length of a royal pendulum, to the length of the pendu- lum required. Example. What is the length of a pendulum that vibrates half- ſeconds, or 120 times in a minute; 120 X 120 = 14400 60 x 60 3600 Then as 14400 : 3600 :: 39.13: 978, &c. inches the length required. Or the length of a pendulum may be found thus. Divide 140848 by the ſquare of the number of vibrations aligned, and the quotient gives the length of the pendulum required. Ехат - 28 P.II. OF PENDULUM S. Example. What is the length of a pendulum that ſwings double ſeconds, or vibrates 30 times ina minute? The fquare of 305900 900) 140868(156.52inches. PROPOSITION I. PROB. Given the length of a pendulum, to find the number of vibrations ſuch a pendulum ſhall make in a minute. RUL E. As the given length in inches : to 39 13 :: fo is the ſquare, of the time given : to the ſquare of the number of vibrations, whoſe ſquare root is the number fought. Examples. How many times does a pendulum 6.78 inches long vibrate in a minute? inch. inch. As 9.78 : 39 13 :: 3600': 144.00, whoſe ſquare root is 120, the number fought. Or divide 140868 by the length of the pendulum, the quotient is the ſquare of the number of vibrations fought. Example. How many vibrations will a pendulum make in a minute, that is 156.52 inches long? inches 156.52)140868(900(30 long. PROBLEM. Holding in my hand a ſtring and bullet, whoſe length from my finger, the point of ſuſpenſion, to the center of the bullet, was 331 inches; I obſerved the faih of a cannon, the ſame inſtant I ſet the bullet a ſwinging and it made i vibrations between the time of ſeeing the flaſh and hearing the report, how far was the cannon from me? inch As 33.5: 39.13 :: 3600 : 4205, whoſe ſquare root is 64.8 the number of vibrations the bullet made in a miuute, then as 64.8: 60 ::11:10.18. But found flies 1142 feet in a ſecond, then 1142 X 10.18 =11625.56 feet = 2.2 miles. PART inch vib. vib. feconds feet [ 29 ] FIG. PART III. OF G Ε Ο Μ Ε G M E T T RY . 1 G 1. DEFINITIONS. EOMETRY is a ſcience of whatever is extended, viz, of lines, fuperficies, and ſolids. It is a frience of enquiring, inventing, and demonſtrating all the affections of magnitude. 2. A point hath neither parts nor dimenſions, and incapable of being divided 3. A line is length without breadth, and is made by the flowing of a point. 4. A right line is the neareſt diſtance between two points as AB 5. A ſuperficies has length and breath, but without thick- neſs. 6. A ſolid hath length, breadth, and depth or thickneſs. 7. An angle is the inclination of two lines the one to the other, meeting in a point called the angular point, as A B, AC, in A. 8. If lines that form the angle be right lines, it is called a right lined angle. 9. When a right line C D ſtanding upon another right 3. line A B, makes on both fides thereof the angles C DA, C D B, equal one to the other, then both thole angles are right angles; and the right line CD, which ſtands upon the other is called a perpendicular. Cor. When an angle is mentioned it is noted by three letters, the middle letter always denotes the angular point, as CDA, D is the angle denoted. 10. An obtuſe angle is greater than a right angle, as EDB; 4. for CDB is a right angle. 2. II. An 30 GEOMETRY. P.III. Fig. II. An acute angle is leſs than a right angle, as ADE; for 4. A D C is a right angle: acute and obtuſe angles are called oblique angles. Cor. The lines that form an angle, are called its legs. 12. An equilateral triangle hath three equal fides, as the triangle A. 13. An iſoſceles triangle hath only two ſides equal, as the triangle B. 14. A right angled triangle hath one right angle as at n, db is called the hypothenuſe, b n the perpendicular, d n the baſe, as the triangle C. 15. The triangle D is called an obtuſe angled triangle, for m is an obtuſe argle. 5 16. The altitude or height of any triangle is the perpendi- 6. cular CD, let fall from the vertex C to the baſe A B; and it may fall either within or without the triangle. 17. Parallel lines are fuch, if infinitely produced, would 7. never meet, as A B, CD. 8. 18. A ſquare A B C D has four equal fides, and is right angled, and therefore equi-angled. 9. 19. A parallelogram ABCD hath its oppoſite ſides parallel, and conſequently equal. IO. 20. A trapezium is a four-fided figure, whoſe oppoſite fides are not parallel; and if a line be drawn acroſs it from corner to corner, it is called a diagonal line, as A B. 21. A circle is a plane figure contained under one line only, called its circumference or periphery. A B is the diameter, AD or CD the radius. Cor. The circumference of every circle is divided into 360 degrees, and is the moſt capacious of any figure. 22. All figures that have more than four ſides are called polygons, and are named from the number of their fides, and when their fides are equal, they are called regular polygons. As if it be of Five fides a pentagon. Six fides an hexagon. Seven fides an heptagon. Eight fides an octogon. Nine fides a nonagon. Ten fides a decagon. Eleven fides endecagon. Twelve fides dodecagon. a II. - 23.A P.III. GEOMETRY. 31 23. A problem is a propoſition which propoſes fomething to Fi&. be done. 24. A theorem is when ſomething is propoſed to be de- monſtrated. 25. A corollary is a conſequence drawn from ſomething that has been already demonſtrated. 26. A fcholium is a remark on any propofition or corollary. 27. An axiom is ſuch a common, plain, ſelf-evident and received notion, that it cannot be made more plain and evi- dent by demonftration, becauſe itſelf is better known than any thing that can be brought to prove it. PRO B. I. To divide a given right line into two equal parts, that is, to biſect a right line. Let the given line be A B, then with one foot of the com- I 2 paſſes in A, opened to any diſtance greater than half the line AB, deſcribe an arch, then with the ſame opening of your compaſſes ſet one foot in B, deſcribe another arch, and where theſe two arches cut one another, as in b and d, if the line b d, be drawn, that will biſect the given line AB. PRO B. H. From a point in a given line to raiſe a perpendicular. Let A B to be the given line, and C the given point; take 13 two points equally diſtant from C, as D, E, and ſetting the foot of the compaſſes in D and E, opened to any diſtance greater than A C, ſtrike two arches, and where they enter fect one another as in b, there draw hc, which shall be the perpendicular required. PRO B. III. At the end of a given line to erect a perpendicular, as ſuppoſe At any convenient diſtance out of the given line as at E, 14. fetone foot of the compaſſes, and extend the other to the given point B, with this diſtance draw a circle, and from the inter- ſection in the given line as at D draw a line through the point E to interſect the arch above in C, then CB drawn will be the perpendicular. N. B. By this problem may as many perpendiculars as you pleaſe be raiſed from given points on a right line, as well as from the end PROB at B a 32 GEOMETRICAL PROBLEMS. P. III. с FIG. PROB. IV. To let fall any perpendicular on any given line AB, from a 15. given point above that line, as ſuppoſe from C. From the point C deſcribe an arch as cd, cutting the line A B, then with the ſame diſtance of the compaſſes and one foot in cand d; deſcribe two arches below, and where they interfeet as at e, draw Ce, it will be the perpendicular required. PRO B. V. With a given right line to make an angle of any number of degrees. Example 1. 16. Let it be required to lay down an angle of 30 degrees 30 minutes on A B from the point A. Take the diſtance of 60 degrees with your compafies from the line of chords, then ſetting one foot in A, with that di- ſtance, deſcribe an arch as CD, then take 30° 36' from the line of chords, and ſet it off from Cro D, then a line drawn from A through D, the angle D AC is the angle required. Example 2. To lay down an angle above 90 degrees. 17 From A, on a given line A B, it is required to make an angle of 145 degrees. With the diſtance of 60 from the line of chords ſweep an arch, then ſince the line of chords extends only to go degrees, fet it off on the arch at twice; as firſt take goº, then 55° to C, and the angle is made by drawing АС. PRO B. VI. Three lines being given to make thereof a triangle. 18. Let the lines be A, B and C, whoſe lengths are 12, 9 and 6 reſpectively. Make A C equal to the line A, or 12 from a ſcale of equal parts, take the length of the line B or 9, and with one foot of the compaſſes in A, deſcribe an arch, then with the compaſſes take the extent of the line C, or 6 equal parts from the ſame ſcale, and with one foot in C deſcribe another arch, and where they cut one another as in B, join AB and BC, and the triangle is conftruated. a PRO B. P. III. GEOMETRICAL PROBLEMS, 33 a FIG. PROB. VII. To bileet or divide an angle A B C into two equal parts. Set one foot of the compaſſes in B, and deſcribe an arch 19. beat any extent, take b d equal to half b c, then through d, draw the line BD, and the angle B will be divided into two equal parts. PRO B. VIII. Through a given point C to draw a parallel line to a given right line AB. Take any point in the given line A B, as F, and with the 20. diſtance F C, deſcribe a ſemicircle ; take the arch E D equal to the arch B C, then through D draw the line CD, and it will be parallel to A B. PRO B. IX. 2. draw a circle through three points, as ABC, not lying in a ſtraight line. Join the points A B, and AC; bifect A B and B C (by 21. prob. 1.) then draw the biſecting lines E and F, and where they meet, as in D, then the point D is the center of the circle required. Cor. By having the ſegment of any circle given, the whole circle may be completed by taking any three points in the ſegment's arch, and proceeding as above. SCOLIU M. By this problem may be known what kind of ordnance an enemy makes uſe of by collecting the ſegments or pieces of the ſhell fhot from their artillery ; for the circle may be per- fected by drawing two lines A B and B C, within the given arch, and biſecting them. PRO B. X. To infcribe in a given circle the ſide of a pentagon. Draw the diameters A G and B H at right angles to one 22. another, biſect the radius B C in F, and draw A F; make FE equal to FC, and through the point E, draw BD; then B D is the fide of the pentagon required. D PRO B. > 34 GEOMETRICAL PROBLEMS. P. III. - FIG. P R 0 B. XI. To inſcribe a circle E F G in a given triangle ABC. 23. Bifeet the angles B and C with the right lines B D, CD, meeting the in point D; and draw the perpendiculars DE, DF, DG, from the center D; deſcribe a circle through E, and it will paſs through F and G, and touch the ſides of the angle. Cor. For the angle DBE = DBF by conſtruction, and the angle DEB = D F B, amd D B common, therefore D E = DF=DG. SCHOLIU M. Hence the ſides of a triangle being known, their fegments which are made by the touchings of the circle inſcribed may be thus found. Let AB=12; AC=18; BC=16. 1. AB+BC--AC=BE+BF=10, and BE=BF=5. And ACEAE+FC: allo FC=CG. Therefore BC-BF=FC=CG=N; and BA-BE=AE -AG. PRO B. XII. To deſcribe a ſegment of a circle upon a given right line AB, that Shall contain an angle equal to a given angle F. Make an angle B A E equal to the given angle F. From G, the middle of A B, raiſe G C perpendicular to A B, and draw A C perpendicular to A E, cutting G C in C; with the radius C A from C, as a center, deſcribe a circle; then will the fegment A D B contain an angle A D B equal to the given angle. PROB. XIII. Upon a given right line AB, to deſcribe any regular polygon. Divide 360 degrees in a circumference, by the number of ſides of the required polygon, and the quotient is the angle at the center; and the angle at the center taken from 180 degrees, leaves the angle at the circumference. Hence is conſtructed the following 24. a TAB L E. P. III. 35 GEOMETRY. Fig. TABLE. Polygons Equal fides Angles at the center. Angles at the circumference. names. 3 4. 5 Trigon Square Pentagon Hexagon Heptagon Octagon Nonagon Decagon Endecagon Dodecagon OOOO arw 120° 90 72 60 51 45 40 oo? oo 00 Oo 25 00 oo oo' oo 00 oo 34 oo 00 } 600 90 108 120 128 135 140 144 147 150 7 8 9 10 II 12 00 oo 36 32 30 4311 164 00 00 The uſe of the above table is to conſtruct any of the regular polygons upon a given right line, as follows. Example 1. Upon a given line A B to deſcribe a hexagon. At A and B make the angles each equal to 120 degrees, 25. as in the table, then draw A C and B D each equal to A B; then at C and D, make angles each equal to 120 degrees as before, and make CE, DF, each equal to AB, draw EF and the hexagon is conſtructed. Example 2. Upon a given line A B to conſtruct a pentagon. Make the angles A and B each equal to 108 degrees as in 26. the table, then draw the lines A C and B D each equal to the given line AB; on the point C and D, with the com- paſſes opened to the diſtance AB, deſcribe arches croſſing each other in E, and the pentagon is conſtructed. By the ſector. But ſuch like queſtions as theſe relating to regular poly- gons may be eaſily anſwered by the ſcales of polygons on the inner ſide of the ſector marked pol. theſe fcales begin at 4 and figured backwards to 12 fides; the tranſverſe diſtance from D 2 6 @ 36 P.III. GEOMETRY. FIG. 6 to 6, (which may be opened to any diſtance) is equal to the radius of the circle circumſcribing the polygon, the legs of the ſector being kept open at the required diſtance; then the interval between 5 and 5, is the ſide of a pentagon ; between 6 and 6, is the fide of a hexagon; between 7 and 7, is the ſide of a heptagon; between 8 and 8, is the fide of an octa- Or if you would have the radius of a circle circumſcribing a regular polygon, open the ſector till the tranfverfe diſtance be equal to the fide of the given polygon, the ſector being kept at that opening, the interval between 6 and 6 is the ra- dius of the circle, in which that polygon may be inſcribed ; this being ſo plain, examples would be needleſs. gon, &c. Some of the moſt uſeful THEOREMS in plane Geometry demonſtrated. I being felf-evident principles. T will be proper to premiſe firſt the following axions, I. Things that are equal to one and the ſame thing, are equal to one another. II. If equal things be added to equal thing, the ſum will be equal. III. If equal things be ſubtracted from equal things, the remainder will be equal. IV. If equal things be multiplied by equal things, the pro. duct will be equal. V. If equal things be divided by equal things, the quotient will be equal. VI. Things which being laid upon one another do agree, or meet in all their parts, are equal to one another. VII. Every whole is equal to all its parts taken together ; and therefore every whole is greater tban its part. VIII. All right angles are equal the one to the other. Τ Η Ε Ο R Ε Μ Ι. If a right line ſtands upon another right line, the angles which it makes with it will either be two right angles, or together equal to two right angles. DEMONSTRATION. 27. Let the lines be AB and CD meeting in the point C, upon which point deſcribe any circle at pleaſure. Then P. III. GEOMETRY 37 Then the arch A D is the meaſure of the LACD, and the FiG. arch D B of the LDCB; but the arches A D + DB = 180 degrees, ſince they complete a femicircle ; conſequently LACD + DCB= 180º = to two right angles. Q: E. D. Corollary 1. Hence it follows, that if ZĀCD= 90°, then muſt DCB = 90°; but if LACD be obtuſe, then the LDCB will be acute. Cor. 2. And if ever ſo many right lines ſtand upon a right line at one and the ſame point, all the angles taken together will be equal to two right angles. Τ Η Ε Ο R Ε Μ ΙΙ. If two lines cut or croſs each other, the two oppofite angles will be equal. DEMONSTRATION. Let the two right lines be A C, DE, croſſing each other 28 in B. Then by theo. 1. 2 ABD + LDBC = 180º = L DBC+ CBE; hence by axiom 3. 2 ABD = 2 CBE. In like manner it is proved that LDBC= L ABE. 2. ED. Cor. Hence it is evident that if two or more right lines interfect each other, in one and the ſame point, they will make twice as many angles as there are lines; and all the angles that are capable of being formed about one point, are equal to a whole circle, or 360 degrees. THEOREM III. Any right line croſſing two parallels makes the alternate angles on the ſame ſide equal to one another, i. e. L AHF = LCGF. DEMONSTRATION. Let the two parallel right lines A B C D be croſſed by 29 the line EF. Since parallel lines are ſuch right lines, in ſome plane, which if infinitely produced will never meet, and Th. parallel lines muſt have the ſame inclination to any right line that crofleth them, and confequently ZAHF=ZCGF. 9. E.D. THEOREM IV. Any right line croſſing two parallels (ſee the laſt figure) makes the alternate angles on different ſides equal, i.e. AHG= LHGD. DEMON- D 3 38 P.III. GEOMETRY. Fig. DEMONSTRATION. By th. 3 11 LAGH=LCGF. 3 I Th.2.2 LCGF= LHGD. By Ax. 1. 132. 1 312AGH = LHGD. 2. E. D. THEOREM V. If a right line croſs two parallels, the two internal angles are equal to two right angles, i.e. LAHG ALCOH= 180°. DEMONSTRATION. By theo. 4. LAHG=LHGD, and by ax. 2. adding 2 > I I=2 AwN Fig. THE OR EM XI. In any right angled triangle BPH, the ſquare HH, deſcribed upon the hypothenuſe H, is equal to the ſum of tbe ſquares BB,PP, deſcribed on the two other fides B, P. DEMONSTRATION. 35 Make a ſquare whoſe fide is equal to B+P, and let that ſquare be repreſented by S, fo will its area be BB+2BP+PP. Alſo the ſum of the areas of the four triangles included in that ſquare, will by menſuration be BP x4=2BP. I ſay, HH = BB+ PP. For 11 SHH+2BP. S and 2 S=BB+2BP+PP. 3 HH+2BP=BB+ 2BP+PP. 3—2BP4 HH=BB+PP. Corollary 1. i H=V BB+ PP. For 2 | HH=BB+ PP. 2002 3H+V BB+ PP. 2. BEV HH-PP. For BB+PPHH. 2-PP 3 BB=HH - PP. 30024 | BEVHH-PP. 3. I PENHH-BB. For PP+BB=HH. 2-BB 3 PP-HH-BB. 30024 PEVAH-BB. = Or this DeMONSTRATION. 36. Let ABC be the right angled triangle, make _ CAD=90°, and draw AD to meet CB produced in D; then by Theo. 19. following BC: BA :: BA : BA2+ BC2 Then will DC- : AC :: AC : BC, and multi- plying extremes and means, we have AC=AB? +BC. 2. E. D. I 2 BA2 = DB. BC BC THEO P. III. GEOMETRY. Ar Τ Η Ε Ο R Ε Μ ΧΙΙ. Fig. In an obtufe angled triangle ABC, if a perpendicular be let fall 36. upon the baſe; or one ſide adjoining to the obtuſe angle B, then the Square of the ſide oppoſite to that obtufe angle is equal to the ſum of the ſquares of the two leſſer ſides, together with twice the rectangle of the baſe and diſtance of the perpendicular from the obtufe angle. DEMONSTRATION. I ſay ACP-AB?+ BC2+2CB x BD. For ACP=CD2+ AD2=CB²+2CBD+BD2=CB2 +2CBD + AB2. Cor. Hence BD-- AC2-CB2-AB? 2CB Τ Η Ε Ο R Ε Μ ΧΙΙΙ. If a perpendicular be let fall upon the baſe, or fide adjoining to 37. the acute angle B, then, The ſquare of the ſide oppoſite to that acute angle, together with twice the rectangle, of the baſe, and the diſtance of the perpendicular from the acute angle; is equal to the ſum of the ſquares of the i wo other ſides : AC?+ 2CB x BD=AB2+ BC2. DEMONSRRATION. For ACP-AD?+DC = ADP + BC2+BD-2BD XBC =AB? + BC2--2CBD. And AC? + 2CBD=AB²+ BC. AB2+BC2-AC? Cor. BD= 2CB 39. > I THEOREM XIV. The diagonal of a parallelogram divides it into two equal triangles. DEMONSTATION. Since AC, BD, and AB, CD, are parallels. By theo. 4. ) 1 ZACB=LCBD. And 2 ZBCD=LABC. Therefore | 31 AABC=ACBD, being equiangular tri- 3 angles, and AC common, and equal to one another in every reſpect. 2. E. D. Τ THEOREM XV. Parallelograms having the ſame baſe and altitude, are equal. DEMONSTRATION. Let the two parallelograms ABFE, CDEF, ſtand upon 40. the fame baſe EF, and between the ſame parallels EF, AD. Then Since GEOMETRY. P. III. FIG Since | AB=EF=CD by conſtruction. Theref. 2ACEBD, and = Becaufe 3 | AE=BF. 3 = And 4 ZA=LB. Then 5 1 A ACE=ABFD, take the triangle BCG from both, and there remains the trapezium ABGE=CDFG; add the triangle EFG to both ſides, and the ſum is ABFE=SCDFE. Q. E. D. Corollary. Since the diagonal of every parallelogram divides it into two equal triangles by theorem 14; therefore triangles conſtituted upon the fame baſe, and between the ſame pa- rallels, are equal. Lemma. A right line is ſaid to be multiplied by another right A Bline, when a right 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 angled parallelogram is made of the two 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 lines; that is, the 17 | 18 | 19 20 | 21 | 22 23 24 product of the two 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 numbers which ex- D C prefle the equal parts in the two fides, are equal to what Mathematicians call the ſuperficial content or area of ſuch figures, or they are underſtood to contain ſo many little ſquares, whoſe fides are each equal to one of the equal parts with which the ſides were meaſured. Thus for inſtance, let AB=CD=8, AD= BC=4, then ADX AB=4 x 8 = 32, which is the area or number of little ſquares contained in the parallelogram, as appears by the figure. THEOREM XVI. Parallelograms, and alſo triangles, that ſtand between the ſame parallels, or which have the ſame height, are in a direct proportion to one another, as their baſes. DEMONSTRATION. By the above lemna. The area of the BF=BE x EF, SBD-BEXDE. which areas are in the proportion of DE: EF. For BEXDE: BEEF:: DE: EF. That is, BD: BF ::DE: EF. SDBE=BE XDE. Again by theo. 14. area A EBF-IBE XEF. Then will A DBE : A EBF ::DE: EF. 2. E. D. 4.1. THE 0. P. III. GEOMETRY. 43 III 11 Id 3 5 THE OR EM XVII. FIG. If a right line be drawn parallel to any ſide of a plane triangle, it will cut from the given triangle another leſs triangle, that will be ſimilar or equiangled to the whole triangle. DEMONSTRATION. By theo. 3. ZB=LC, ZE=2D, and ſince LA is com- 42. mon, it is evident that the triangle ABE is fimilar or equi- angled to the triangle ACD. 2. E. D. THE OR EM XVIII. If a perpendicular be let fall from the right angle of a right an- 43. gled triangle upon the hypothenuſe, it will divide the triangle into two right angled ones, ſimilar to the whole and to each other. DEMONSTRATION. Draw the lines as by the figure where _CDB=LBDA= LCBA=90°; therefore _CBD+_DBA=90°, and by cor. theo. 6. 2C+ZCBD=90°, hence by ax. 1. 2CBD+L DBA=LC+ZCBD, ſubtract LCBD and there remains DBA=LC In like manner ZA+Z ABD=LABD+ZDBC=90°, and ſubtracting ZDBA, we have ZA=_DBC: hence it is proved that CLDBA, and ZA= ¿DBC therefore the triangles CDB, BDA, CBA are fimilar. Q. E. D. SCHOLIU M. Whence we have the following proportions from the ſimilar triangles CDB, BDA, CBA. AC: CB :: AB: BD. CB: BD:: AB: AD. CB: BD :: AC: AB. AB: BD :: CB: DC. DC: BD :: BD: AD. AD: AB :: AB: A C. CD: CB :: CB: AC. CD: BD :: CB: A B. AD:BD :: AB : CB. Τ Η Ε Ο. 44 GEOMETRY. P.III. FIG. THEOREM XIX. If two triangles be ſimilar, their like fides, or the fides fubtending equal angles will be proportional. DEMONSTBATION. 44. Let DE be parallel to CB, then will the triangles ABC, AED be equiangular, join DB and CE, then by cor. to Theo. 15. | AECD=ADBE. By theo. 16. A ADE: AABD :: AE : AB AndAADE : ACE :: AD: AC. Since the ADBE= AECD, therefore A ABD=AACE. Hence AE: AB :: AD : AC. 2. E. D, THEOREM XX. If any angle of a plane triangle be biſected by a right line, it will cut the oppoſite ſide in proportion to the other two ſides of the triangle. DEMONSTRATION. 45. Produce the fide DC till CZ=CB, join ZB, and draw CF parallel to BD, then will the triangles CFZ, DCA be fimilar, for By theo. 3 ZZCFELD. And th. 8. 2 Z+_ZBC=_BCD. But 31 Z=LZBC. Becauſe 4 | ZC=BC. Therefore 5 LZ=LACD. And 6 ZA=LF. Since 71 2 BCA=_FBC, the lines CA, ZB, are parallel, and Thence 8 | BA=FC, and 9 BA (=FC): BC(=ZC) :: AD: CD or which is the Same 10 AB : AD :: BC : CD. 2. E. D. Τ Η Ε Ο R Ε Μ ΧΧΙ. . If a right line be divided into two parts or ſegments, the ſquare of the whole line will be equal to the ſquare of each ſegment, together with a double rečiangle of the ſame ſegments. DEMONSTRATION. Let CG be the ſquare of AC and GE that of AB, and let HE, DE be produced to meet the ſides of the ſquare GC in B and 1 Theo. 19. 1.0 a 46. P. III. GEOMETRY. 45 a 3 > ) B and F, then fince DE=EH, EC will be the ſquare of BC, Fig. and AE, -FH= a rectangle under A B, BC, but CG=GE + EC+ AE+FH; that is, AC=AB 2 + BC2+2AB X BC. Or thus, Let the given right line AC be cut any wife as in B, then the ſquare deſcribed on the whole line AC is equal to the ſquares deſcribed on the ſegments AB, BC, and to twice a rectangle made of the fegments AB, BC, together. For AC X AB=AB 2 + AB X BC. And AC X CB=CB 2 + CB XAB. Theref. AC=AB:+CB 2 + 2 AB X CB, In numbers, Let AC=10, AB=6, and BC= 4, Then AC XAB=10x6 .. = 60, AB² = 361 =60, AB X BC = 245 Allo ACXCB=10X4 - - =40, = 162 CB8AB=24 -40, Theref. AC=ABP+ CB12+2 AB x BC=100. 2. E.D. Τ Η Ε Ο R Ε Μ ΧΧΙΙ. The angle at the center of any circle is double to the angle at the periphery, when both the angles ſtand upon the ſame arch. DEMONSTRATION. 2 - CB 2 Caſe I. Let the diameter AD and line BD form the angle at 47. the periphery, draw the radius CB, then I ſay, ZAČB is double ZD. For | BCA=2D + 2DBC, by th. 8. But | _D=LDBC, by th. 9. becauſe DC=CB. Theref. | LBCA=22D. Cafe II. 48. I Draw the Diameter AD, then by Cafe I. 1 ZACB=2LADB. And LACE=2 LADE. I +231 LACB+ LACE=2LADB+2 LADE, by 2 ax. 2. That is, 4 LECB=2_EDB. Caſe 46 P. III. GEOMETRY. I ax. 3 I FIG. Caſe III. Draw the diameter AD, then by 49. Caſe I. ZACB=2LADB. And 2 ZACE=2LADE. 1-21 31 LACB— LACE = 2 LADB — 2ADE, by That is, 4 LECB=2LEDB. Corollary. Hence all the angles at the periphery that ſtand upon the ſame or equal arches are equal to one another. THEOREM XXIII. The angle in a ſemicircle is a right angle. DEMONSTRATION. 50. Let ABC be the diameter, and draw the radius BD, then by Th. 9. LALADB. And 21 LCLCDB. = 1+2 13 | ZA+ZC=LADB+ LCDB, byax. 2. That is 4 IZA+ZC=LADC. But 5 | ZA+ZC+ ZADC=180°, and Theref. 6 | LADC=90°. Q.E.D. THE OR EM XXIV. An angle in a ſegment leſs than a ſemicircle is greater than a right angle, and an angle in å ſegment greater than a ſemicircle is leſs than a right angle. DEMONSTRATION. 51. Let the LABC be an angle in the ſegment AEBC greater than a ſemicircle; and LADE an angle in the ſegment CDA leſs than a ſemicircle. Draw the diameter CE, and join BE, ED, then by theo. 23. LEDC = 90°= LEBC, it is thence evident that ŁADC is greater than _EDC, and alſo that LABC is leſs than _EBC. THEOREM XXV. Τ If two lines be any how drawn within a circle cutting each other, the rectangle of the ſegments of one line will be equal to the rectangle of the ſegments of the other line. DEMONSTRATION. 52 Join the points a, b, and c, d, then will the triangles aob, and cod be fimilar, for La=LC, Lb=ld, by cor. to theo. 20, therefore by theo. 19, 00:00:20: ob. And by multi- plying extremes and means we have co xob=ao xod. 9. E. D. Τ Η Ε Ο. a P. III. GEOMETRY. 47 by Fig. THE ORE M XXVI. If from a point without a circle two right lines be drawn, to the oppoſite part of the periphery; the rectangle of one whole line and its part without the circle is equal to the rectangle of the other whole line, and its part without the circle. DEMONSTRATION. Draw the lines BC, AD, then the triangles VBC, VAD 53. are fimilar, for 2D=2B by cor. tu theo. 22, and Required the number anſwering to the log. 6.4213713? OPERATION. Next the greater logarithm Next leſs logarithm 6.4212748 Difference 1646 6.42 14394 - Given logarithm 2638000.0 586.2 6.4.213713 6.4212748 2638586.2, anſwer. 1646).965.0000(586.2 8230 I 4200 13168 &c. MULTIPLICATION by logarithms. CASE I. To find the product of two given whole or mixed numbers. RULE. Find the logarithm of each given number, and their fum will be the logarithm of the product, whoſe correſponding number had from the tables, is the anſwer. Example 1. Multiply 84 logarithm 1.92427 by 25 1.39794 Product 2100 3.32221 Example 2. Múltiply 41.5 logarithm 1.61804 by 7.24 0.85973 Produét 300.4 2 47777 CASE II. When both or either of the factors are leſs than unity. RULE. K 2 13% OF LOGARITHM S. P. VII, RULE. Remove the decimal point towards the right hand till the firſt figure becomes a whole number, with which take out the logarithms, and find the product as by cafe I. then remove the decimal point as many places towards the left hand as you before removed it to the right, and it is the true product. Example 1. reinoved. Multiply 54 - 5.4 logarithm 0 73239 by .25 2.5 0.32794 - Product 1:35 1 3:5 I.13033 Example 2. removed. Multiply .74 7-4 logarithm o 86923 by 5.25 -5.25 0.72016 Product 3.885 38.85 1.58939 Example 3. removed. Multiply :742 - 7:42 logarithm 0.87049 by -421 4.21 0.62428 - Product .312382 31.2382 1.49468 Division by logarithms. CASE I. To divide a whole or mixed number by a leſs whole or mixed number. RUL E. From the logarithm of the dividend fubtract the logarithm of the diviſor, and the remainder is the logarithm of the guotient. Example 1. Divide – 624 logarithm 2.79518 Ву 26 1.41497 Quotient 24 4:38021 Example P. Vit. OF LOGARITHMS. 133 Example 2. Divide I 221 logarithm=3.08671 by 1.91062 81.4 Quotient = 15 =1.17609 DAN Example 3 Divide- 34:86 logarithm=1.54237 8.3 20.91907 by Quotient = 4.2 =0.62325 C À SE II. When both or either of the factors are leſs than unity: RUL E. Remove the decimal points till the factors contain whole numbers, and the dividend the greateſt, with which find the product as before. Then if the dividend be more places removed than the dia viſor, by the exceſs remove the quotient to the left hando But if the diviſor be more places removed, by the exceſs res move the quotient to the right, and it is done. Example 1. removed. Divide 92.42 -- 72.42 logarithm=1.85983 by •543 5.43 =0.73480 Quotient 1.33370 1.33370 I.12505 - Example 2. removed. Divide 87.5 logarithm = 1.94200 by 51.43 - 51.43 1.71121 Quotient .17013 1.7013 1.7613 = =0.23079 875 - 1995 - Divide -7824 64321 Qyo. 1.81069 by Example 3 removed. 7.824 logarithm=6.89342 4.321 =0.63558 1.8106 0:25784 PROP K 3 134 P. VII. OF LOGARITHMS. PRO P. XV. The Square, cube, &c. any given number by logarithms. RUL E. Multiply the logarithm of the given number, by the index of the power fought. Note, 2 is the index of the ſquare, or ſecond power. 3 is the index of the cube or third power. &C. & C. E c. Example 1. Example 2. What is the ſquare of 51? What is the cube of 12 ? 51 logarithm=1.70757 1 2 logarithm=1.07918 Index Index 3 Anſwer 260 1 =3.41514 Cube is 1728 = 3.23754 . 2 PRO P. VI. To extract the roots in logarithms. RULE. This is only the reverſe of the former rule, namely, divide the logarithm of the given number by the propoſed index, the number anſwering to the quotient is the required root. Example 1. Example 2. What is the ſquare root What is the cube root of of 144? 1728? 144 logarithm= |2.15836 | 1728 logarithm= [ .23754 Anfwer 12=1.01918 Anſwer 12 =1.07918 2 3 To work the rule of three direct logarithmically. Add the logarithm of the ſecond and third terms together, prom which fum take the logarithm of the firſt term, and the remainder is the logarithm of the fourth term fought, where adius is 10. As is evident to any one that purſues the ſolu- tions to the following caſes in trigonometry. PART [ 135 ] Ρ PART VIII. OF PLANE TRIGONOMETRY. a TA RIGONOMETRY is a rule by which we learn'how to compute the length of the fides and quantity of the angles of all manner of triangles, from proper data. In order to which there is calculated and diſpofed into tables, the length of the fines, tangents, and ſecants of each degree and minute of the quadrant of a circle whoſe radius is 10000000000, and theſe are called natural fines, tangents and ſecants. But with regard for accommodating the aforeſaid tables to practice, the logarithms of the natural fines, tangents, and ſecants are diſpoſed into tables, which are called artificial fines, tangents, and fecants. And it is not only requiſite that the circumference of a circle be divided into degrees and minutes, but likewiſe certain lines both within and without the circle are to be expreſſed by parts of the radius. DEFINITIONS. 1. CB-radius of the circle AGIBE. 2. IB=chord, a right line joining the extremities of the the arch IB together. 3. IF=chord of the arches IBF, and IAF. 4 ID=fine of the arches IB and IA. 5. CD=co-fine of the arch IB. 6 BT=tangent of the arches IB and IA. 7. CT=ſecant of the arches IB and IA. 8. DB=verſed line of the arch IB, and 9. DA==verſed fine of the arch IA. 10. The complement of an arch is what it wants of a qua- drant, or go degrees; then GI is the complement of BI, for BG is a quadrant. 11. GH=co-tangent of the arch IB. 12.CH=co-ſe cant of the arch IB. 13. GK=co-verſed fine of the arch IB. CO. K 4 136 PLANE TRIGONOMETRY. P. VIII. :: : : fine : fine : co-fine : ſine : :: : : : : COROLLARY. Hence the co-fine, co-tangent, co-ſecant, co-verſed fine of an arch, is equal to the fine, tangent, fecant, verſed fine of its complement reſpectively. SCHOLI U M. As co fine radius tangent. radius fecant tangent. radius fecant radius. radius co-ſecant : radius. tangent : radius radius co-tangent. co-tangent : co-fine co fecant : radius. fine ; co-fine tangent radius. fine : co-fine radius ço-tangent. co-fine : radius radius fecanr. fine : radius radius CO-ſecant. ſecant : tangent radius fine. fecant : radius :: radius co-fine. Hence 1. Radius ſquare=fine ſquare + co-fine ſquare= ſecant ſquare --tangent fquare=tangent multiplied by the Co-tangent. 2. Sine=V..dius 1quare-C0-1ine square. 3. Co-line =V rauiusiquaie-- Muelquare. five fecant 4. Tangent= C0-lecant tangent Co-ſecant 5. Secant fire =V rad.19.+ co-tangent tangent tquare. 6. Vericd fine=radius-co-fine. rad. fquare 7. Co-tangent tangent 8. Co-fecant tine The foregoing properties of fines, tangents, &c. are the natural Gnes, natural tangents, &c. and they muſt be actually multiplied, and divided, and here radius is always I. Example What is the co-fecant of 15° 01'? The natural fide of 15° or is =.2591, and radius=i, and conſequently its ſquare 1. Then .2591)1.0000000(3.859;1, the co-ſecant required; and fo of the ref. But co Gne rad. fq. CO-110e rad. ſquare P. VIII. PLANE TRIGONOMETRY. 137 But what here follows, belongs to the logarithmic fines, logarithmic tangents, &c. and the radius here is always 10: and then addition and ſubtraction performs the work. Thus, 1. Log. radius + log fine-log. co-fine=log, tangent. 2. Twice log, radius - log. co-line=log. fecant. 3. Twice log. radius-log. tangent=log. tangent of the complement of that arch to a quadrant. Example. 1. What is the log. tangent of 30° 10'? The log. radius 10.0000000 + log. fine of 38° 10' 97909541 Sum 19-7909541 Take log. cofine, 38° 10' 9 8955422 Log. tangent of 38° 10' 9.8954119 Example 2. What is the log. ſecant of 38° 10'? 2 log. radius 20. -log. cofine 38° 10' 9.8955422 Log. fecant of 38° 10' 10.1044578 Example 3. What is the log.co-tangent of 38° 10'? 2 log. radius 20. -log. tangent 98954119 Log. co-tangent of 38° 10' 10.1045881 PRO P. I. PROB. To find the fine, co-fine, tangent, &c. to go degrees, of the in- termediate parts of a minute. RUL E. Take the difference between the fines, &c. of the given degrees and minutes, and of the minute next greater. Then, as 1 : is to that difference :: fo is the given intermediate pare of a minute in decimals : to a fourth number. Therefore multiply that difference by ſuch decimal part, and add the product to the fine, tangent, and fecant; or, ſubtract it from the 138 PLANE TRIGONOMETRY. P. VIII. the co-fine, co-tangent and co-ſecant of the given degrees and minutes, the ſum or remainder, will be the fine, tangent, &c. required. Example 1 What is the natural fine of 1° 48' 28" 12''? The nat. fine of 1° 49' 317015 nat. fine of 1° 314108 Difference 2907 Now as 1 : 2907 :: (28'' 12'''=)0.47 : 1366 The natural fine of 1° 48' 3'4108 481 The nat. fine of 1° 48' 28" 12" 3:5474 Example 2. What is the log. fine of 1° 48' 28" 12"'? The log, ſine of 1° 49' 8.5010798 log. fine of 1° 487 8.4970784 Difference, 40014 Now 28'' 12'' =0 47 Then 40014 X.473 18805 Add log. fine of 1° 48' 8.4970784 Log. fine 1! 48' 28'' 12'' = 8 4989589 - Example 3. To find the log, co-fine of 88° 11' 31' 48'"? Log. co fine of 889 IT' 8.5010798 Log. co-fine of 88 12 8.4970784 Difference 40014 But 31' 48'"=0.53 Then 40014 x 0.53 From co fine 889 II' 8.5010798 Take 21207 Log. co-fine 88° 11' 31' 48'" = 8.4979591 21207 cm Example P. VIII. 139 PLANE TRIGONOMETRY. 8.6223427 8.6193127 2 Example 4. Required the log, tangent of 2° 23 33" 36"? Log. tangent of 20 24 Rog. tangent of 2 23 Difference 30300 But 33' 36'' =0' 56 To log. tangent of 2° 23' 8.6193127 add 30300 X.56= 16968 Log. tangent 2° 23' 33" 36"= 8.6210095 ' '= PRO P. II. Given the natural or logarithmic fine, co-fine, tangent, &c. to find the arc. o RUL E. Take the next lefs of the ſame kind found in the tables, and fubtract it from that given, obſerving the degrees and minutes anſwering to it; then annex 2 or 3 cyphers to the remainder; divide it by the difference between it and the next greater, the quotient will be the decimal part of a mi- nute, to be added or ſubtracted from the degrees and minutes before found. Example 1. Given this log. fine 9.8393859 to find the arc? Firſt, the given log. fine is= 9.8393859 The next leis is 43° 41' = 9.8392719 Difference 1140 The diff between the 43° 41' = 9.8392719 And the next greater 43 42 9.8394041 IS 1322 Then 1322)1140 000(0.862 Anſwer 43° 41'.862=43° 41'51" 43'". Example 2. Given the log, tangent 9.6766687 to find the arc ? Firſt the given log. tangent is 9.6766687 Next leſs is log. tangent 25° 24' 9.6765426 Difference 1261 The 140 PLANE TRIGONOMETRY. P.vill. The difference between it and the next greater, viz. 25° 25' is 3260, then 3260)1 261.000(0'.387 And 25° 24.387=25° 24' 23' 13", anſwer. Example 3. Given log. co-fine. 9.9994206 to find the arc? Firſt log. co-fine given 2.9994206 Log. co-fine of 2° 58' next leſs 9.9994176 Difference 30 Diff. between 2° 58' and 2° 59'=65 Then 65)30.0000(0.4615 From 2° Take 0.4615 581 2 57-5385 Or, 2° 57' 32" 19", anſwer. PRO P. III. PROB. Given the radius and arch of a circle in degrees and minutes, 18 find the length of thar arch. RU L E. Multiply the radius of that arch by .017453, and this pro- duct by the number of degrees and decimal parts, gives the length of the arch, in the ſame meaſure the radius is of. Example What is the length of an arch of 106° 28' whoſe radius is 24 yards ? .01745 24 radius 6980 3490 41880 106.46=106° 28' 251280 167520 251280 418800 44.5854480 yards, anſwer. This P. VII. PLANE TRIGONOMETRY. 141 This rule is uſeful for finding the meaſure or length of the circular part before a baſtion, of the ditch before a ravelin, of the retired Hank, &c. PRO P. IV. PROB. Given the ſum of two angles, and the product of their fines, to find the angles ſeparately. RU L E. Divide the product of the fines, by .5, and to the quotient add the co-fine of the ſum of the angles, and the ſum will be the co-fine of the difference of the angles; which difference added to, or ſubtracted from the ſum, the ſum or difference is double the greater, or double the leſs angle reſpectively. Example 1. Suppoſe the product of the fines of two angles be .294114, and their ſum 8 19 30', what are the two angles ? •5).2941141 Add cofine 81° 30' = .58822 .14780 .73602 co-fine of the difference Sum of the angles=42° 36. To 81° 30' Add 42 36 From 81° 30 Take 42 36 2)}24 06 2) 38 54 62 3 1927 Anſwer greater angle=620 3" { leſs angle 19 27 Example 2. Suppofe the product of the ſides of two angles be .44131, and their ſum=84° 32', what are the two angles? :5).441316 •88262 Add co-fine .09526 of 8432' Co-line= .97788 of the difference of the angles=12°4'. То 142 PLANE TRIGONOMETRY. P. VIH. FIG. To 840 32 From 84° 32' Add 12 Take 12 4 4 2)96 36 2)72 28 48 18 36 14 Anſwer S greater angle=48° 18' = 36 14 {fels angle 93. AXIOMS I. In right angled triangles, any fide being made the radius, 92 femidiameter or ſweep of a circle, the other two will be fines, tangents or ſecants, which words being writ upon, or ex- pounded by them, it will be As the word on the given fide, Is to the given fide; So is the word on the fide required, To the ſide required. N. B. The word on any ſide of a triangle, is to be under- 94. ſtood the fine, tangent or fecant of the degrees and minutes which are contained in the angle expreſſed by that word. II. The fides of any plane triangle, whether right or oblique angled, are in proportion to one another, as the fines of their oppofite angles. III. In any plane triangle, As the ſum of any two fides Is to their difference So is the tangent of half the ſum of their oppoſite angles To the tangent of half their difference. IV. In any plane triangle As the baſe is to the ſum of the two other fides, ſo is the difference of thoſe fides to the difference of the ſegments of the baſe made by letting fall a perpendicular from the angle oppoſite the baſe to the baſe. V. Half the ſum of any two numbers being added to half their difference, gives the greater number. But, Half the difference taken from the half ſum, gives the leſs number. VI. Having the three ſides of an oblique triangle given, to find an angle. Set P. VIII. PLANE TRIGONOMETRY. 143 1 Fig. Set down the ſide oppoſite to the angle required, and under it the other two fides; from their half ſum ſubtract each fide in the order they ſtand, then find the complement arithmeti. cal of the logarithms of the half ſum, and firſt remainder, and alſo the logarithms of the other two remainders, then will the half ſum of theſe four logarithms be the tangent of half the angle fought. The arith.com. of a logarithm is found by beginning at the index, and writing down what each figure wants of 9, all but the laſt, which muſt be taken from 10. When there are two figures in the index, reject the firſt, and work as before. VII. Any two fides of a plane triangle taken together, are greater than the third that remains. VIII. The greateſt fide of every triangle is oppoſite to the greateſt angle, and the greateſt angle oppoſite the greateſt fide. IX. The ſum of the three angles of every plane triangle is equal to 180°; therefore X. If two angles of a plane triangle are known, the third is alſo known, being found by taking their ſum from 180 degrees. XI. If any angle be obtuſe or greater than 90 degrees, each of the other two will be acute or leſs than go degrees. XII. If one angle be right or go degrees, the ſum of the other two will be go degrees. XIII. If one of the acute angles, in a right angled plane triangle, be given, the other is alſo known, being found by taking the given angle from goº, and this defect is called its complement. CASE I. The hypothenuſe and angle at the baſe being given, to find the perpendicular. It will be proper to obſerve, 1. When a fide is required, begin with an angle oppofite to a given fide. 2. When an angle is required, begin the proportion with a fide oppofite to a given angle. Example. In the triangle ABC, right angled at B. Are given SAC=34 2 LA=28° 4' Required AB and BC. Geometrical 95 14.4 PLANE TRIGONOMETRY. P. VIII. Meaſure {BC=16] Geometrical ſolution. Draw any right line AB, make ZA=28° 4' and draw the hypothenufe AC=34 from any ſcale of equal parts, fitted to your line of chords, from the point C, let tall the perpendicu. lar BC, and it is done. 30 on the ſame ſcale of equal parts. = Numerical ſolution. 7. By the natural fines, Hypothenufe AC radius. From 90° 0 Take LA= 28 4 LC= 61 56 by ax. XIII. The natural fine of 28° 4.47049 natural fine of 61 565.88240 Radius=1. Hypothenuſe AC=34. rad. AC nat. f. LA. As : 34 :: .47049 34 . 188196 141147 15.99666 or 16, very nearly, the perpendicu- lar BC. rad. AC nat. f. LC : 34 :: .8824 34 Ass 35296 26472 30.0016=AB. 10.000000 2. By logarithms. As radius, fine 9(ZB) To hypot. AC, 34 its log. So is fine LA=28° 4 To BC=16 its log. 15.31479 9.672558 1.204037 To P. VIII. TRIGONOMETRY. 145 FIG To find AB As rad. fine 90° 10.000000 Is to AC=34 log: 1.535479 So is fine Z.C=61° 56' 9.945066 To AB= 30 log. 1.377145 Gub II. Baſe AB radius. By the natural tangents, &c. sec. LA AC As 1.13327 : 34 :: 5332 34 1 tang. ZA 21328 15996 1.13327)18.1288(16-BC fec. LA AC rad. As 1.13327 : 34 :: I: 1.13327)3400000(30=AB By logarithms. As fecant LA=28° 4' co. ar. 9.945666 TO AC=34 1.531479 So is tangent LA 2804 9.726892 To BC=16 As ſecant LA=280 04 1.204037 10.054334 To AC = 34 So is radius, fine 900 1.531479 10.000000 TO AB=30 1.477145 IIF. Perpendicular BC radius, By natural ſecants and tangents. nat. ſec. LC AC rad. BC As 2.1254 34 :: I: 16 {ec. LC AC tang. LC AB 2.1254 : 34 :: 34 :: 1.87545.: 30 L 11 146 P. VIII. TRIGONOMETRY. FIG. By the logarithms. As fecant C=61° 56' 10.327442 Is to AC=34 So is the radius, fine 90° 1.531479 10.000000 To BC=16 1.204037 As fecant _C=61° 56' co. ar. 9.672558 TO AC=34 1.531479 So is tangent 2C=61° 56' 10.273107 To AB=30 1.477144 Inſtrumentally. The extent from 90° to 28° 4' on the fines, will reach from 34 to 16=BC on the numbers. The extent from 90 to 61° 56' on the fines reaches from 34 to 30=AB on the numbers. CASE II. Given the angle at the baſe and the perpendicular, to find the hypothenuſe and baſe. Example. 99. In the triangle ABC right angled at B S BC=12 =° Geometrical ſolution. At any point B draw BC perpendicular to AB, which make =12 from any ſcale of equal parts; then from the line of chords, make the LC=53° 8' (for 90° - 36° 52'=53° 8) draw the hypothenuſe CA and ic is done. are given {LAE1360 52'} required AC and AB. Meaſure SAC=20 ZAB=16 I ſhall not any more give the operation by the natural fines, natural tangents, &c. but leave it to the exerciſe of the learner; which is very proper he ſhould underſtand. Numerical ſolution. I. Hypochenuſe AC radius. From Take LA 3652 96. 900 2C= 53 8 P. VIIL. TRIGONOMETRY. 147 As fine ZA= 36° 52' 9.778119 FIG. To BC=12 So is radius 90° 1.079181 10.000000 TO AC=20 1.301062 As fine ZA=36° 52' co. ar. 0.221881 To BC=12 1.079181 So is fine _C=53° 8' 9.903108 TO AB=16 1.204170 98, II. Bare AB radius. 97: As tangent LA=36° 52' co. ar. 0.1 24990 To BC=12 1.079181 So is ſecant LA=36° 52' 10.096892 To AC-20 1.301063 As tangent ZA=35° 52' 9.875010 To BC=12 1.079181 So is radius, tangent 45° 10.000000 To AB=16 1.204171 III. Perpendicular BC radius. As radius, tangent 45° 10.000000 To BC=12 1.079181 So is ſecant LC=53° 8' 10.221881 TO AC=20 1.301062 As radius, tangent 45° I0.000000 To BC=12 1.079181 So is tangent 2C=530 8 1.0124990 To AC-26 1.204171 Inſtrumentally. The extent from 36° 52' to 90° on the fines, will reach from 12 to 20=AC on the number, the extent from 369 52' to 53° 8' on the fines reacheth from 12 to 16=AB on the numbers. L2 CASE 348 TRIGONOMETRY. P.VIII FIG. CASE III. The two acute angles and bafe being given, to find the hypothenuſe and perpendicular. Example. 100. Let the baſe AB=36, and the angle A=36° 52"; required the hypothenuſe AC, and perpendicular BC. Geometrical ſolution. Draw the baſe AB, which makes = 36 from a ſcale of equal parts, raiſe the perpendicular BC, make the angle A=36% 52', drawing the hypothenuſe AC to cut the perpendicular BC in C, and it is done. Meafure SAC=45 96. > BC=27 Numerical ſolution. I. Hypothenuſe AC radius. As fine LC 53° 8 9.903108 To radius, fine goº 10.000000 So is AB= 36 log. 1.556302 Co, ar. 900 . To hypoth. AC=45 1.653194 As fine _C=53° 8' co. ar. 0.096892 Is to AB=36 1.556302 So is fine LA=36° 52' 9.778119 97. To perpendicular BC=27 1.431313 II. Baſe AB radius. As radius, fine 90° 10.OCOCOO To AB= 36 log. 1.556302 So is fecant A=36° 52 10.096892 To hypoth. AC=45 1.653194 - As radius 90° 10.000000 Is to AB=36 1.556302 So is tangent LA=36° 52' 9.875010 To perpend. BC=27 1.431312 III. VIII. TRIGONOMETRY. 149 . 98. Fie III. Perpendicular BC radius. As tangent 2C=53° 8' co. ar. 9.875010 To AB= 36 1.556302 So is fecant 2C=53° 8' 10.221881 1.653193 10.1 24990 To hypoth. AC=45 As tangent 2C=530 8 Is to AB = 36 So is the radius, 90° 1.556302 10.000000 0 98. To perpendicular BC=27 1.431312 Inſtrumentally. The extent from 53° 8' to 90° on the fines, will reach from 36 to AC=45 on the numbers. The extent from 53° 8' to 362 52' on the fines, will reach from 36 to BC=27 on the numbers. CAS E IV. The hypothenuſe and perpendicu.ar being given, to find the baſe and angles. Example. Let the hypothenuſe AC=40, perpendicular BC=24; required the baſe AB, and angles. Geometrical ſolution. Draw AB, upon which raiſe the perpendicular BC=24, then with the extent of 40 and one fooc in C, cut BA in A, join AC, and it is done. =32 = 36° 52' 90 00 Remains LC= Numerical ſolution. 1. Hypothenuſe AC radius. As AC=40 1.602060 To radius 90° 10.000000 So is BC= 24 1.380211 To fine LA=36° 52' 9.778151 90 00 45 53 08 Meaſure { LA 53 08 96. I 3 As 150 TRIGONOMETRY. P. VIII. FIG. As radius 90° 10.000000 Is to AC=40 So is fine ZC=53° 08' 1.602060 9.903108 To AB= 32 1.505168 II. Perpendicular BC radius, As BC= 24 1.330211 98, Is to radius 90° So is AC=40 10.000000 1.602060 To fecant 2C=53° 8' 10.221849 90 O 36 52 LA As radius, 909 10.000000 To BC= 24 1.380211 So is tangent 2C=53° 8' 10.124.990 To AB=32 1.505201 Inſtrumentally. The extent from 40 to 24 on the numbers, will reach from 90 to LA=36° 52' on the fines. The extent from 90° to 53° 8' on the fines, will reach from 40 to AB=32 on the numbers, CASE V. Given the hypothenuſe and baſe, to find the perpendicular and the angles. Example 99. Let the hypothenuſe AC=73, baſe AB=55, required the perpendicular and angles. Geometrical ſolution. Draw the baſe AB=55, at B erect the perpendicular BC, then with the extent of 73 and one foot in A, cut BC in C, draw the hypothenuſe AC, and it is done. SBC =48 2 4 Ion Meaſure LA 90 00 LC- 48 53 Nume- P. VIII. TRIGONOMETRY. 157 Fro. 96, Numerical folution. I. Hypothenufe AC radius. As AC=73 1.863323 To radius, 90° 10.000000 S is AB=55 1.740363 To fine ZC=48° 53' 9 877040 90 00 LA= 41 07 As radius, fine 90° I0.000000 To AC=73 So is fine 2A=419 07' 1.863323 2.817958 To BC=48 1.681281 97 II. Baſe AB radius. 1.740263 As AB=55 To radius, 900 So is AC=73 I0.000000 1.863323 To fecant LA=41° 07' 10.122960 90 00 ZCE 48 53 As radius, 90° 10.000000 90° To AB=55 1.740363 So is tangent LA=41° 07' 9.940946 To BC=48 1.681312 Inftrumental ſolution. The extent from 73 to 55 on the numbers, will reach from to 2C=48° 53' on the fines. The extent from goº to 41° 07' on the fines, will reach from 73 to 48 =BC on the numbers. CASE VI. Given the baſe and perpendicular to find the hypothenuſe and angles. Example I 4 152 TRIGONOMETRY. P. VIII. Fig. Example. 99 Let the baſe AB=45, perpendicular BC=28, required the hypothenuſe AC and angles, Geometrical ſolution. From any ſcale of equal parts, make AB=45, raiſe the perpendicular BC, which make equal to 28 from the ſame Icale, join the points A and C, and it is done, Meaſure ŞAC=53 31° 53' 90 00 ZC- 58 07 Numerical ſolution. I. Baſe AB radius. 1.653212 97. As AB=45 To radius, tangent 450 So is BC=28 10.000000 1.447158 To tangent ZA=31° 53' 9793946 As radius, tangent 45° 10.000000 To AB=45 1.653212 So is ſecant ZA=319 53 10.071028 TO AC=53 1.724240 II. Perpendicular BC radius. As BC=28 1.447158 98. To radius, tangent 45° 10.000000 So is AB=45 1.653212 To tangent 2C=589 7' 10.206054 das As radius, tangent 45° 10.000000 To BC=28 1.447158 So is fecant LC=58° 7' 10.277209 = TO AC=53 1.724367 I:firu. P. VIII. 153 TRIGONOMETRY. Q Q Inſirumentally. Fig. The extent from 45 to 28 on the numbers, will reach from 4:0 10 31° 53 = LA on the tangents. The extent from 31° 53 to 90°, on the fines, reaches from 28 to AC=53 on the numbers. Obſervation. By making the hypothenuſe radius, each fide becomes fines of their oppoſite angles, and therefore the ſides are in proportion to the fines of their oppofite angles, and the contrary by, axiom 2, and this muſt be allowed the moſt elegant when it can be uſed; which I have done in every cale except the laſt. OBLIQUE PLANE TRIANGLES. CASE I. Given one ſide and the angles to find the other two ſides. Example. 101. In the oblique triangle ACB LA=28° 4 are given B=53 8 required AC and BC LAB=42 Geometrical conſiruction. Make AB=42, draw AC making the angle" A=28° 4', alſo making the LB=53° 8', draw BC, and where it meets AC, the triangle is conſtructed. Calculation. From 180° oo Take SLA=280 : 4 4 IZB=53 : 8 Sum= 81 12 Remains LC= which uſe. 98 48 its ſupplement is 81° 12' As 154 TRIGONOME TRY. P. VIII. FIG. As fine 2C=90° 48' 999485 JOJ. Is to AB=42 1.62325 So is fine ZA=28° 4' 9.67255 To BC=20 1.30095 As fine LA=28° 4' 9.67255 Is to BC=20 1.30095 So is fine B=530 87 9.90310 TO AC=34 1.53150 Or you may find the ſides thus, As fine _C=91° 48' co. ar. 0.00514 Is to AB=42 1.62325 So is the fine ZA=28° 4' 9.67255 To BC=20 111.30094 As fine _C=98° 48' co. ar. 0.00514 Is to AB=42 1.62325 So is fine _B=53° 8 9.90310 TO AC=34 1.53149 CASE II. Two ſides and one angle oppuſite to them being given, to find the reſt. N. B. When the given angle is oppoſite to the greater given ſide, this caſe has one certain anſwer: but when the given angle is oppoſite to the leſs given fide, this caſe will then have two true anſwers, as will evidently appear by the following examples. Example 1. In the oblique triangle ACB (AC=85 are given BC=50 JO1. ZB=53° 8. required AB and L. Cand A. Geometrical conſtruction. Draw AB, and alſo BC=50 making B=53° 8', then with the extent of 85, and one foot in C, cut BA in A, and it is done. Meaſured P.VIII. 155 TRIGONOMETRY. Meaſured {AR=105 280 4 1.92941 As AC=85 To fine ZB=530 8 So is BC=50 9.90310 1.69897 To fine LA=28° 4' 9.67266 From 180° oo Take SLA=28° 4 ELB=53 8 Sum =81 12 Remains ZC=98 48 by ax. 10. 9 90310 As fine _B=53° 8 Is to AC=85 So is fine _C=98° 48' 1.92941 9.99485 TO AB=105 2.02116 Or thus. As AC=85 co. ar. 8.07058 To fine ZB=53° 8' 9.90310 So is BC=50 1.69897 To fine ZA=28 4 9.67265 ZB=53 8 81 12 Take From 180 00 LC=98 48 As fine ZB=53° 8' co. ar. 0.09689 Is to AC=85 1.92942 So is fine _C=98° 9.99485 484 TO AB=105 2.02119 Example 156 TRIGONOMETRY P. VIII. Fig. Example 2. Let} AC=85 required AB and L. B and C? SBC59 3 C LA=2804 Geometrically. Draw AB, and AC=85, making the LA=28° 4, then with the extent of 50, and one foot in C, cut AB, in the two points B, b, join BC, 6C, and either of the triangles will be that required. Meaſure { 2.ACB:5,8° 48' ZAC6=25° 4'. Ab=45. Calculation. As BC=50 co. ar. 8.30103 Is to the fine 2 A=28° 4 9.67255 So is AC=85 1.92942 ioz, To fine _B=LCUB=53° 8' 9.903 From 180°-00 Take {LASS4 =5308 =284 Sum 81 12 Remains LACB 98 48 Fron 2CbB=53° 8' Take LA 28 4 Remains ZACb= 25 4 And LA C=126 52 by ax. 10. As fine LA=28° 4' co. ar. 0.32744 Is to BC=50 1.69897 So is fine ZACB=98° 48' 999485 To AB=105 2.02126 As fine ZA=28° 4' co. ar. 0.32744 Is to BC=50 1.69897 So is Gine LACb=25° 4 9 62703 TO AL=45 1.65344 CASE P. VIII. TRIGONOMETRY. 157 FIG CASE III. Two ſides and their contained angle being given, to find ot the reſt. Example. AB=64 Let {AC=48 required BC and LLC and B. LLA=360 14 Geometrically. IOI. Make AB=64, and LA=36° 14', draw AC which make 48, join BC, then will the triangle ACB be that required, BC meaſures 38, and LB=48° 18'. Calculation by axiom 3. AB=64 From 180° oo' AC=48 take LA= 36 14 Sum fides=112 Sum=143 46=LL B and C. Half ſum 71 53 As fum fides 1 1 2 co, ar. 7.95078 Is to their diff. 16 1.20412 So is tang. half LZSB & C=71° 53' 10.48522 To tangent half their diff.=23° 35' 9 64012 To half ſum=71° 53 Add half diff. =23 35 From half fum=719 53 Take half diff. =23 35 Greater LC=95 28 Leſs LB =48 18 As fine ZB=48° 18' Is to AC=48 So is fine LA=36° 14 cor. ar. 0.12689 1.68124 9.77164 To BC= 38 1.57977 CASE IV. The three ſides being given, to find the angles. Example 158 TRIGONOMETRY P. VIII. AB=522 FIG. Example. 103 Let 3 BC=32 required LLLA, C, and B? AC=40 Geometrically. Make AB=52, with the extent of 40, and one foot in A, ſtrike an arch alſo, with the extent of 32, and one foot in B, cut the former arch in C, join AC and BC, and it is done Make the perpendicular CĎ. Meaſures LA=37° 57' {LB=50 15 Calculation by axiom 4. AC=40 AC=40 BC=32 BC=32 diff=8 Sum=72 Bafe. fum, diff. AS 52 : 72 :: 8:11.07=diff. ſegments of the baſe. 5:53=half diff. feg, To half baſe =26 Add half diff. feg.= 5:53 Sum=greateſt ſeg. =31.53=AD. From half bafe =26 Take half diff. fegm = 5.53 Remains leſs feg. =DB20.47 SAC hyp40 Given AD the bale= 31.53 } to find the LsA and ACD. As AC=40 1.60206 10.00000 1.49872 Is to radius, fine 90 So is AD = 31.53 To fine LACD=52° or' From 90 00 9 89666 Remains LA=3759 As P. VIII. TRIGONOMETRY. 159 As BC=32 1.50515 F.o. 10.00000 1.31111 Is to radius, fine 90° So is BD=20.47 To fine _BCD=39° 46' From 90 00 9.80596 Remains 2B=50 14 Now ZCELACD+ ZBCD=91° 47'. For ACD=52° 01' And BCD=39 46 91 47 Otherwiſe by axiom 6. BC=32 AB=52 AC=40 124 Half ſum=62 30 Remainders 10 log, 22 co, ar. 8.20760 co, ar, 8.52288 1.00000 1.34243 19.07291 Half is tangent of 18° 581 9.53645 2 - LA=37 57 as above, hence the reſt will be found by axiom 2. Or thus. 103 AB2 + BC2-AC2 A theorem =co-fine B. 2 XAB X BC ABP + AC2-BC2 And theorem -=co-line of the LA. 2ACX AB To find the angle B. In numbers. Here AB=52, its ſquare=2704 BC=32, its ſquare=1024 AC=40, its ſquare=1600 And 160 TRIGONOMETRY. P. VIII. TIG. And AB x BC=52X32=1664 X2=3328 2704 1024 3728 1600 3328)2128.00000(-63942 the natural co-line of 50°15'=2B. To find the angle A. 52 X 40 = 2080 X 2=4160 Then 2704 1600 4304 1024 1 4160)3280.00000(78846 the natural co-fine of 37° 58'=LA. Whence their fum taken from 180°, leaves the angle C= 91° 47. I ſhall now proceed to ſhew the various uſes of a right angled triangle, and the oblique, with regard to the taking of heights and diſtances, &c. PROBLEM I. To find the height of a tower, caſtle, church, &c. when you can come to the foot of it. 304 With any inſtrument to take angles, as ſuppoſe a quadranty you fland ſome diſtance from the foot of the tower, as at A, and there taking your quadrant, and directing the fights of it to C, and find the ſtring cuts the limb at 480 12', then find by meaſuring in a ſtraight line how far you ſtood from the foot of the tower, and note the meaſure, fuppoſe here 45 yards, then may she height of the tower be found. BC denotes the tower or perpendicular of the right angled triangle ABC AB the diſtance meaſured 45 yards, or baſe of the triangle, AC the viſual fight from the quadrant to the tower's top, or hypothenuſe. Now P. VIIT 161 TRIGONOMETRY. Now there is given LA=480 : 12 FIG. 2C=41 : 48 to find BC. AB=55 yards By Caſe III. As fine ZC=41° 48' 9.82382 Is to AB=45 1.65321 So is LA=48° 12' 9.87243 To BC=50.33 1.70182 The height of the inſtrument above ground muſt always be added, which here ſuppoſe was two yards above ground, then 52.33 yards is the tower's height. PRO B. II. To find the breadth of a river. 104. An engineer comes to the ſide of a river, as at B, where he is to lay a bridge over it to paſs the army, and wants to know the breadth of it, in order to know how many pontons he muſt uſe upon that occaſion. Standing at B, he ſees ſome object, ſuppoſe at C, on the other ſide of the river, and then making a right angle with his inſtrument, he obferves another object by the river fide at A, then meaſuring to it finds it 49 yards, and the L A=53° 20', hence the breadth of the river may be found thus. Given LA=530 20 LC=36 40 to find BC. AB=49 yards As radius 10.00000 a Is to AB, 49 So is tangent LA, 53° 20' 1 69019 10.12815 To BC=65.82 1.81834 But when the caſe happens you cannot come to the foot of a tower, &c. by reaſon of water, enemies, or other obſtacle, it is then called an inacceffible height, and may be found by the following M PROB 162 P. VIII. TRIGONOMETRY. a FIG. PROB. III. To meaſure an inacceſſible height. 105 At any convenient diſtance from the tower, as at A, find the angle of altitude, fuppofe 28° 34', then meaſure in a ſtraight line towards the tower, as to D, and ſuppoſe it mea- fures from A to D 30 yards, then at D take another angle of altitude, which let be 50° 9'. Here the fight lines AC, DC, and the diſtance line AD form an oblique triangle ADC, wherein are given the diſtance AD=30, and all the angles to find DC, and conſequently BC. Firſt CAB=28° 34' LBDC=509 9 ' _DCB=39° 51' by ax. 13. But LADC=129° 51'=LDCB+ 2B=39° 51'+90® by theorem 8. or 180°-50° g'=129° 51' by theorem 1. Then A 28° 44' LADC=129º 51 AD=30 Take 158 25 from 180 oo Remains 2 ACD=21 35 by ax. 10. Or thus. LCDB=500 g From Take LA =28 34 Remains LACD=2135 As fine LACD=21° 35' 9.56567 Is to AD=30 So is fine ZA=28° 34' 1.47712 9.67959 To fight line DC=39 1.59104 As radius 10.00000 Is to hyp. DC=39 So is fine _D=5099 1.59104 9.88520 1.47624 To BC=30 ferè, altitude N.B. P. VIII 163 TRIGONOMETRY. FIG. N. B. You may make uſe of this problem for finding the length of unapproachable lines, as thoſe of places beſieged; but never make the angle DCA leſs than two degrees. PROB. IV. 106, To find the height of a. caſtle ſtanding upon a hill, and your diſtance from it. Firſt find the CAD ſuppoſe=40°, and GAB=22°, then meaſure in a ſtraight line towards the caſtle, as from A to D, let be 40 yards; then at D find LCDB, let=63° 20', and LGDB = 49°, and you may from theſe data find the height, and diſtance DB. Solution. ¿ CAD=400 LGAB=220 LCDB=630 20 GDB= 49° _DCB=26° 40' by ax. 13. =90 LADC=116° 40' =_DCB+ LB, by theorem 8. ZACD=23° 20' by ex. 10. And AD 40 yards. As fine LACD=23° 20' 9.59778 ZB Is to AD=40 So is fine _CAD=40° 1.60206 9.80806 To fight line CD=64.91 1.81234 As radius 10.00000 Is to CD=64.91 So is fine CDB=63° 20' To BC=58 the hill and caſtle 1.81234 9.95115 1.76349 As radius 10.00000 Is to CD=64.91 So is fine 2 DCB=26° 40' 1.81234 9 65205 To DB=29.13 1.46439 M 2 As 164 P. VIII. TRIGONOMETRY. FIG. As radius 10.00000 To DB=29.13 1.46439 So is the tangent LGDB=49° 10.06083 To BG=33.51 the alt. hill 1.52522 From altitude of hill and caſtle BC=58 yards. Take BG the perpend. hill 33:51 Remains CG the Caftle's height=24.49 PRO B. V. 107. Caſe I. Let A, B, C be three churches whoſe diſtances are, viz. AC=106, BC=65.5, AB=53.25, and ſuppoſe at ſome fta- tion, as at E, I can ſee all thoſe objects, and would know their reſpective diſtances from me, I take an angle AEB=139 30', alloZCEB=29° 50', from theſe data their diſtance from me may be found as follows, Through A, C, and E deſcribe a circle, and draw AE, CE, and BE, continue BE to D, and join AD, DC, then there are given AC=106 BC=65.5 ZAEB=13° 30'= LAED. _CEB=29° 50' =CED. LAEC=43° 20'= LAEB + 2CEB. LADC=136° 40'=180°-LAEC, by theo. 8. ZACD=13° 30'=LAED?