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COMPLETE KEY 
 
 UMMERE'S SURYEYmG; 
 
 THE OPERATIONS OF ALL THE EXAMPLES NOT SOLVED IN 
 THAT WORK ARE EXHIBITED AT LARGE. 
 
 PRINCIPALLY DESIGNED 
 
 TO FACILITATE THE LABOUR OF TEACHERS, 
 
 AND TO ASSIST THOSE 
 
 WHO HAVE NOT THE OPPOETUNITY OF THEIR INSTRUCTION. 
 
 / 
 By SAMUEL ALSOP. 
 
 ADAPTED. TO THE REVISED EDITION OF THE SURVEYING 
 By ISAAC SHARPLESS, 
 
 Author of "A Text-book of Geometry and Trigonombtky." 
 
 PHILADELPHIA: 
 
 PORTER & COATES. 
 
 \ 
 
Copyright, 1884, 
 BY PORTER & COATES. 
 
 
 '<Pf7 
 
 L^X)(r.C: 
 
KEY 
 
 TO 
 
 GUMMERE'S SURVEYING. 
 
 PLANE TRIGONOMETRY. 
 
 CASE 1. 
 
 Example 3. (PI. 1, fig. 1.) 
 
 Angle C=180°— A— B=46° 15'. 
 
 As sin. A 19° 23' - - Ar. Co. 0.007499 
 
 Is to sin. B 54° 22' - - - 9.909963 
 
 So is EC 125 ■ 2.096910 
 
 To AC 103.4 2.014372 
 
 Again, 
 
 As sin. A - Ar. Co. 0.007499 
 
 Is to sin. C 46" 15' 9.858756 
 
 SoisBC 2.096910 
 
 To AB 91.87 1.968165 
 
 Example 4. (PI. 1, fig. 2.) 
 Angle C=90°— A=33° 12'. 
 
 As sin. C 33° 1:2' Ar. Co. 0.261566 
 
 Is to sin. B 90 - - - - 10.000000 
 
 So is AB 53.66 1.729651 
 
 To' AC 98 - - 1.991217 
 
 As sin. C - - - ^ Ar. Co. 0.261566 
 
 Is to sin. A 56° 48'^ 9.922603 
 
 So is AB - - - 1.729651 
 
 To BC 82 ----- 1.913820 
 
PLANE TRIGONOMETRY. [Case 2. 
 
 Example 5. (PL 1, fig 2.) 
 Angle C=90°— A=50° 50'. 
 
 As sin. A 39° 10' - Ar. Co. 0.199573 
 
 Ts to sin. B 90° - - 10.000000 
 
 So is BC 407,37 2.609989 
 
 To AC 645 2.809562 
 
 As sin. A Ar. Co. 0.199573 
 
 Is to sin. C 50° 50' - 9.889477 
 
 SoisBC 2.609989 
 
 To AB 500.1 - - -' 2.699039 
 
 ::;ase 2. 
 
 Example 3. (PI. 1, fig. 1.) 
 
 As AC 306 Ar. Co. 7.514278 
 
 Is to AB 274 2.437751 
 
 So is sin. B • 78° 13' 9.990750 
 
 To sin. C - - 61 14 - 9-942779 
 
 139 27 
 180 
 
 A - - 40 33 
 
 As sin. B 78° 3' - Ar. Co. 0.009515 
 
 Is to sin. A 4113^ 23' -9.812988 
 
 So is AC 306 2.485721 
 
 To BC 203.4 2.308224 
 
 Example 4, (PI l,fig. 2.) 
 
 As AC 272 - - - - Ar. Co. 7.565431 
 
 •Is to AB 232 2.365488 
 
 So is sin. B - 90° lOOOOMO 
 
 To sin. C . - 58° 32' - 9.930919 
 
 A - - 31° 28' 
 
 / 
 
Cases.] PLANE TRIGONOMETRY 
 
 As sin. B Ar. Co. 0.000000 
 
 Is to sin. A 31° 28' 9.717673 
 
 So is AC - - - 2.434569 
 
 ToiBC 142 2.152242 
 
 Example 5. (PI. 1, fig. 2.) 
 
 As AC 150 Ar. Co. 7.823909 
 
 Is to BC 69 - - 1.838849 
 
 So is sin. B - 90° 10.000000 
 
 To sin. A - 27° 23' 9.662758 
 
 C - 62° 37' 
 
 As sin. B Ar. Co. 0.000000 
 
 Is to sin. C 62° 37' 9.948388 
 
 So is AC 2.176091 
 
 ToAB 133.2 - - 2.124479 
 
 CASE 3. 
 
 Example 2. (PL 1, fig. 3.) 
 
 A + C^180°-B^3^. 15' 
 2 2 
 
 As AB+BC 185 Ar. Co. 7.732828 
 
 IstoAB— BC33 - - 1.518514 
 
 So is tang. -^±^ - - 39° 15' - - - - - 9.912240 
 
 ° 2 
 
 To tang. 5ll^ - - 8° 18' 9.163582 
 
 ^ 2 = 
 
 C - 47° 33' 
 
 As- sin. A 30° 57' ------ - Ar. Co. 0.288792 
 
 Is to sin. B 101° 30' 9.991193 
 
 So is BC 76 - - - - 1.880814 
 
 To AC 144.8 2.160799 
 
PLANE TRIGONOMETRY. [Case 4. 
 
 Example 3. (PL 1, fig. 2.) 
 
 :45° 
 
 C+A_180°— B_._^ 
 
 a 2 
 
 As AB+BC 1677 Ar. Co. 6.775467 
 
 Is to AB— BC 103 2.012837 
 
 So is tariff. ?±A - - 45° 10.000000 
 
 ^ 2 
 
 To tang. ^^ - - 3° 31' 8.788304 
 
 C - - 48° 31' ~ 
 
 As sin. A 41° 29' Ar. Co. 0.178878 
 
 Is to sin. B 90 10.000000 
 
 So is BC 787 2.895975 
 
 To AC 1188 3.074853 
 
Case 4.] 
 
 RIGHT-ANGLED TRIANGLES, 
 
 
 
 CASE 4. 
 
 
 
 
 Rule 2. 
 
 
 
 Example 2. (Pi. : 
 
 flg. 4.) 
 
 AC 
 
 47 
 
 
 
 AB 
 
 64 
 
 Ar. Co. 
 
 8.193820 
 
 BC 
 
 34 
 2)145 
 
 Ar. Co. 
 
 8.468521 
 
 Half sum 
 
 72.5 
 
 - -. - - 
 
 - 1.860838 
 
 Difference 
 
 25.5 
 
 - - - - 
 
 ■ - 1.406540 
 
 
 2)19.929219 
 
 ( 
 
 :os. i B 
 
 B 
 
 EXAMP 
 
 22° 49' - - 
 
 . - 9.964609 
 
 
 45° 38' 
 
 
 
 LE 3. (PI. 1, 
 
 fig. 4.) 
 
 AB 
 
 108 
 
 
 
 BC 
 
 54 
 
 Ar. Co. 
 
 8.267606 
 
 AC 
 
 88 
 2)250 
 
 Ai. Co. 
 
 8.055517 
 
 
 IS"! 
 
 
 - - 2 096910 
 
 
 17 
 
 
 - . 1.230449 
 
 
 
 
 
 2)19.650482 
 
 Co 
 
 s.iC 
 C 
 
 48° 2' - 
 
 =. . 9.825241 
 
 
 96° 4' 
 
 
 
 
 
 RIGHT ANGLED TRIANGLES. 
 
 First Method. 
 
 Example 2. (PL 1, fig. 5.) 
 Making AC radius, CB is sine of A, and AB is cos, A ; hence, 
 
8 PLANE TRIGONOMETRY. [Case 4. 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. A 27° 46' 9.668267 
 
 So is AC 36.57 1.563125 
 
 ToBC 17.04 - -' 1-231392 
 
 And, 
 
 As radius - - - - Ar. Co. 0.000000 
 
 Is to COS. A 9.946871 
 
 So is AC 1.563125 
 
 To AB 32.36 - - - 1.509996 
 
 Example 3. (PI. 1, fig. 5.) 
 
 Making AC radius, we have CB the sine, and AB the cosine ot 
 A; hence, 
 
 As sin. A 42° 9' Ar. Co. 0.173230 
 
 Is to radius 10.000000 
 
 So is BC 193.6 2.286905 
 
 To AC 288.5 2.46013 5 
 
 And, 
 
 As sin. A Ar. Co. 0.173230 
 
 Is to cos. A - 9.870047 
 
 So is BC - - - - 2.286905 
 
 To AB 213.9 2.330182 
 
 Example 4. (PI. 1, fig. 6.) 
 
 Making the base AB radius, we have BC the tangent of A ; making 
 AC the radius, we have AB the cosine of A ; hence, 
 
 As AB 46.72 Ar. Co. 8.330497 
 
 Is to BC 57.9 1.762679 
 
 So is rad. 10.000000 
 
 To tang. A 51° 6' ----- 10.093176 
 
 And', 
 
 As COS. A Ar. Co. 0.202066 
 
 Is to rad. 10.000000 
 
 SoisAB - - 1.669503 
 
 To AC 74.4 1.871569 
 
 / 
 
HEIGHTS AND DISTANCES. 
 
 Second Method. — By Logarithms. 
 Example 3. 
 
 Hypothenuse 
 Base - - - 
 
 403 
 321 
 
 
 Sum - - - - 
 
 724 - - 
 
 . log. 2.859739 
 
 Difference - - 
 
 82 - - 
 
 - -" 1.913814 
 
 
 2)4.773553 
 
 Perpendicular 
 
 243.65 - 
 
 - - - 2.386776 
 
 Example 4. 
 Perpendicular 27.2 - - - log. 1.434569 
 
 Base - 
 
 - 31.04 - - 
 
 2.869138 
 - . 1.491922 1.491922 
 
 23.835 - - 
 
 - - 1.377216 
 
 54.875 
 
 1.739374 
 
 3 41.27 - - 
 
 2)3.231296 
 1.615648 
 
 
 
 APPLICATION OF PLANE TRIGONOMETRY TO THE 
 MENSURATION OF DISTANCES AND HEIGHTS. 
 
 Example 1. (See fig. 54, Surveying.) 
 
 Angle C=180°— A— B=56° 23'. 
 
 To find AC: 
 
 As sin. C 56° 23' Ar. Co. 0.079480 
 
 Is" to sin. B 49° 23' - 9.880289 
 
 So is AB 500 yards 2.698970 
 
 To AC 455.8 --.... 2.658739 
 
10 PLANE TRIGONOMETRY. 
 
 To find BC : 
 
 As sin. C Ar. Co. 0.079480 
 
 Is to sin. A 74° 14' - . . . 9.983345 
 
 So is AB - - - - 2.698970 
 
 To BC 577.8 2.761795 
 
 Example 2. (Fig. 55, Surveying.) 
 
 As BC + AC 1575 Ar. Co. 6.802719 
 
 ~ Is to BC— AC 105 - - - 2.021189 
 
 A+B 
 
 So is tang. — ^ 62° 10' 10.277379 
 
 To tang. -^^=?- 7° 12' 9.101287 
 
 B - - 54° 58' 
 
 As sin. B Ar. Co. 0.086818 
 
 Is to sin. C 55° 40' 9.916859 
 
 So is AC 735 2.866287 
 
 To AB 741.2 2.869959 
 
 Example 3. (Fig. 56, Surveying.) 
 Angle CAD=180— ADC— ACD=31° 10' 
 
 To find AC : 
 
 As sin. CAD 31° 10' ----- - Ar. Co. 0.286065 
 
 Is to sin. ADC 53° 30' 9.905179 
 
 So is CD 300 - - - ' 2.477121 
 
 To AC 465.98 ----- 2.668365 
 
 Angle CBD= 180— BCD— BDC = 22° 55' 
 
 To find CB: 
 
 As sin. CBD 22° 55' Ar. Co. 0.409613 
 
 Is to sin. CDB 98° 45' 9.994916 
 
 So is CD - 2.477121 
 
 To CB 761.47 •. . . . 2.881650 
 
HEIGHTS AND DISTANCES. 11 
 
 To find AB: 
 
 As BC + AC 1227.45 Ar. Co. 6.910995 
 
 Is to BC— AC 295.49 2.470542 
 
 So is tariff, —-?±^^ - 71° 30' - - - - 10.475480 
 ^ 2 
 
 * To tang. ^AB— CBA . 350 44' ... . 9.857017 
 ^ 2 = 
 
 CBA - 35° 46 
 
 As sin CBA ...-.---- Ar. Co. 0.233226 
 
 Is to sin. BCA 37° . - . 9.779463 
 
 So is CA 465.98 - - - 2.668367 
 
 To AB 479.8 ----- 2.681056 
 
 Example 4. (Pig. 57, Surveying.) 
 
 To find C : 
 
 AB 3 
 
 AC 2 Ar. Co. 9.698970 
 
 BC 1.8 - ... - Ar. Co. 9.744727 
 
 2)6.8 
 
 Half sum 3.4 0.531479 
 
 Diflference .4 -. —1.602060 
 
 2)19.577236 
 
 Cos. iC' - 52° 4' ----- - 9.788618 
 
 C - 104° 8' 
 
 To find BD: 
 
 As -sin. D 17° 47' Ar. Co. 0.515105 
 
 Is to sin. C 104° 8' ^ 9.986651 
 
 So is BC 1.8 0.255273 
 
 ToBD 5.715 - - - 0.757029 
 
12 PLANE TRIGONOMETRY. 
 
 To find CD : 
 
 As sin. D „ . . Ar. Co 0.515105 
 
 Is to sin. DBC 58° 5' - - 9.928815 
 
 So is BC - - - ■ 0.255273 
 
 To CD 5,003 - - 0.699193 
 
 Example 5. (Fig. 58, Surveying,) 
 
 To find BAC: 
 
 BC 7.2 
 
 AB 12 - - . - Ar. Co. 8.920819 
 
 AC 8 ~ - - - Ar. Co. 9.096910 
 
 27.2 
 
 Half sum 13.6 . = ..... 1.133539 
 
 Difference 6.4 .-.--.. 0.806180 
 
 2)19.957448 
 
 Cos. h BAC 17° 47'i .... 9.978724 
 
 BAC 35° 35 
 
 To find AE: 
 
 As sin. AEB 136° ...... Ar. Co. 0.158229 
 
 Is to sin. EBA 19° ........ . 9.512642 
 
 So is AB 12 .... - 1.079181 
 
 To AE 5.624 -....--.... 0.750052 
 
 To find ACE: 
 
 As AC + AE 13.624 ..... Ar. Co. 8.865696 
 Is to AC— AE 2.376 - - - 0.375846 
 
 So is tang. ^^^ + ^^^ . 84° 42'i - . - 11.033291 
 To tang. AEC— ACE _ g^ ij . „ . 10.274833 
 
 AEC . 146 44 
 
 ACE - 22° 41' 
 
 / 
 
HEIGHTS AND DISTANCES. 
 
 To find AD: 
 
 As sin. ADC 19° Ar. Co. 0.487358 
 
 Is to sin. ACD 22= 41' 9.586179 
 
 So is AC 8 0.903090 
 
 - To AD 9.476 0.976627 
 
 To find CD: 
 
 As sin. ADC Ar. Co. 0.487358 
 
 Is to sin. DAC 138° 19' ....... - 9.822830 
 
 So is AC - - - . 0.903090 
 
 To DC 16.34 1.213278 
 
 To find BD : 
 
 As sin. ADB 44° ....... Ar. Co. 0.158229 
 
 Is to sin. BAD 102° 44' -..-..- - 9.989186 
 
 So is AB 12 - - - 1.079181 
 
 To DB 16.85 1.226596 
 
 13 
 
 Example 6. (Fig. 59, Surveying.) 
 
 To find ABC : 
 
 
 
 
 AC 46 
 
 
 
 
 AB 50 
 
 
 Ar.Co. 
 
 8.301030 
 
 BC 40 
 
 
 Ar. Co. 
 
 8.397940 
 
 2)136 
 
 
 
 
 Half sum "' 68 
 
 - 
 
 - . - . 
 
 - 1.832509 
 
 Difference 22 
 
 30° 
 
 8' - - 
 
 . 1.342423 
 
 
 2)19.873902 
 
 Cos. ^ ABC ^ 
 
 - 9.936951 
 
 ABC 1 
 
 60° 
 
 16' 
 
 
14 PLANE TEIGONOMETRY. 
 
 To find CD and CE : 
 
 As sin. ADC 60° 16' Ar. Co. 0.061309 
 
 Is to radius 10.000000 
 
 So is AC 46 .. - 1.662758 
 
 To CD - - 52.98 1.724067 
 
 CE=iCD = 26.49 
 
 Also CAE=90°— ADC=90°— ABC=29° 44'. 
 
 Example 7. (Fig. 61, Surveying.) 
 Making DE radius, EC is tangent of D ; hence, 
 
 As radius Ar. Co. 0.000000 
 
 Is to tang. D 47° 30' 10.037948 
 
 So is DE 100 2.000000 
 
 To EC - 109.13 2.037948 
 
 EB - 5 :^= 
 
 EC . 114.13 
 
 Example 8. (Fig. 62, Surveying.) 
 
 To find DC: 
 
 As sin. ACD 25° Ar. Co. 0.374052 
 
 Is to sin. CAD 26° 30' ----■... . 9.649527 
 
 So is AD 75 ft. -----...- . 1.875061 
 
 To DC 79.18 ft. 1.898640 
 
 To find BC: 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. CDB 51° 30' . 9.893544 
 
 So is CD - 1.898640 
 
 To CB 61.97 ft. ..-..-... . 1.792184 
 
 Example 9. (Fig. 63, Surveying.) 
 To find DC: 
 
 As sin. ACD 23° 50' . ... Ar. Co. 0.398535 
 
 Is to sin. CAD 44° 9.841771 
 
 So is AD 134 - - - - . - 2.127105 
 
 To DC 230.4 - - - 2.362411 
 
 / 
 
HEIGHTS AND DISTANCES. 15 
 
 To find CE: 
 
 As sin. CED 141° Ar. Co. 0.201128 
 
 Is to sin. CDE 16° 50' 9.461782 
 
 So is CD - - 2.362411 
 
 To CE 106 2.025321 
 
 To find EB: 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. CDB 67° 50' 9.966653 
 
 So is CD 2.362411 
 
 To CB - - 213.3 - - .- 2.329064 
 
 CE - - 106 = 
 
 BE - - 107.3 
 
 Example 10. (Fig. 64, Surveying.) 
 
 To find BD: 
 
 As sin. CDB 17° 15' ..... Ar. Co. 0.527914 
 
 Is to sin. BCD 23° 45' 9.605032 
 
 So is CB 60 - - - 1-'^7^1^1 
 
 To BD 81.49 1-911097 
 
 To find AD: 
 
 As BD + BA 121.49 ..... Ar. Co. 7.915460 
 
 Is to BD— BA 41.49 - - - - - - . . 1.617943 
 
 So is tang. BAD+BDA _ ggo ^q> ... io.427262 
 
 ^ 2 
 
 To tang. BAD— BDA _ ^^o 35' . - - 9.960665 
 BDA - 27° 5' 
 
 As sin. ADB 27° 5' Ar. Co. 0.341716 
 
 Is to sin. ABD 41° .----.-- - 9.816943 
 
 So is AB 40 1.602060 
 
 To AD 57.64 1.760719 
 
16 PLANE TRIGONOMETRY. 
 
 Example 11. (Fig. 65, Surveying.) 
 
 To find AD: 
 
 As sin. CAD 27° Ar. Co. 0.342953 
 
 Is to sin. ACD 138° - - - 9.825511 
 
 So is CD 132 ----- - 2.120574 
 
 To AD 2.289038 
 
 To find AB; 
 
 As sin. ABD 109° Ar. Co. 0.024330 
 
 Is to sin. ADB 8° 9.143555 
 
 So is AD 2.289038 
 
 To AB 28.64 1.456923 
 
PRACTICAL QUESTIONS. 17 
 
 PRACTICAL QUESTIONS. 
 
 Example 1. (PI. 1, fig. 2.) 
 Making AB radius, BC is tangent of A. 
 
 As radius - - - - Ar. Co. 0.000000 
 
 Is to tang. A 52° 30' - - - 10.115020 
 
 So is AB 85 - ... 1 1.929419 
 
 To BC 110.8 2.044439 
 
 Example 2. (PI. 1, fig. 2.) 
 
 Make AB radius, then will BC be the tangent of A ; making AC 
 radius, AB will be the cosine of A ; hence, 
 
 As radius - - - Ar. Co. 0.000000 
 
 Is to tang. A 61° 45' - - - 10.269767 
 
 So is AB 73 - 1.863323 
 
 To BC 135.9 - - - 2.133090 
 
 And, 
 
 As cosine A .- Ar. Co. 0.324845 
 
 Is to radius ----.:..------- 10.000000 
 
 So is AB 1.863323 
 
 To AC 154.2 2.188168 
 
 Example 3. (PI. 1, fig. 7.) 
 
 To find BD. We have in the triangle ABD, the angles and side 
 AB. Hence, 
 
 As sin. ADB 31° Ar. Co. 0.288161 
 
 Is to sin. BAD 100° ^ 9.993351 
 
 So is AB 339 2.530200 
 
 To BD 648.2 - - - 2.811712 
 
 2 
 
18 PLANE TRIGONOMETRY. 
 
 Again, in ABC we have the angles, and side AB, to find BC. Thus, 
 
 As sin. ACB 22° 30' Ar. Co. 0.417160 
 
 Is to sin. BAC 36° 30' 9.774388 
 
 So is AB 339 - : 2.530200 
 
 To BC 526.9 2.721748 
 
 In DBC we have the sides DB and BC, and included angle DBC= 
 72°. To find the side DC. Thus 
 
 BCD + BDC = 1 80°— 72° =108°. 
 Then, 
 
 As BD + BC 1175.1 Ar. Co. 6.929925 
 
 Is to BD— BC 121.3 2.083861 
 
 So is tang. ^CD + BDC^ _ ^^^ 10.138739 
 
 ^ 2 
 
 To tang. BCD— BDC . go 5, 9.152525 
 
 BDC 45° 55' 
 
 And, 
 
 As sin. BDC 45° 55' Ar. Co. 0.143677 
 
 Is to sin. DBC 72° 9.978206 
 
 So is BC 2.721748 
 
 To CD 697.64 2.843631 
 
 This example might have been solved by finding AD =496.76, 
 AC=759.33; whence the angle ADC would be found to be 76° 55', 
 and CD =697.64, as before. 
 
 ExAMPiifl 4. (PI. 1, fig. 8.) 
 Construction. 
 
 With the given distances construct the triangle ABC. Make ACE 
 and CAE respectively equal to 13° 30' and 29° 50'. About the 
 triangle AEC describe the circle ACD. Join EB, and produce it to 
 meet the circumference in D, which will be the situation of the 
 Dbserver. 
 
 Since the angles ADE and ACE are' subtended by the same arc, 
 we have ADE=ACE=13° 30'. Also CDE=CAE=29° 50'. 
 
PRACTICAL QUESTIONS. 19 
 
 Calculation. 
 
 In the triangle ABC, we have the three sides to find the angle 
 BAG. Thus, 
 
 BC 262 
 
 AC 404 Ar. Co. 7.393619 
 
 AB 213 Ar. Co. 7.671620 
 
 2)879 
 
 Half sum 439.5 ...... 2.642959 
 
 DifFerence 177.5 ..... 2.249198 
 
 2)19.957396 
 Cos. i BAC 17° 48' - - - 9.978698 
 BAC 35° 36' 
 
 In the triangle ACE we have the angles and side AC, to find AE. 
 
 As sin. AEC 136° 40' Ar. Co. 0.163523 
 
 Is to sin. ACE 13° 30' - - - 9.368185 
 
 So is AC 404 ... - - 2.606381 
 
 To AE 137.43 .--.....--'. 2.138089 
 
 In the triangle ABE we have the sides AB and AE, and the 
 included angle B AE = B AC + C AE = 65° 26'. To find ABE, thus : 
 As AB + AE .350.43 ------ Ar. Co. 7.455399 
 
 Is to AB— AE 75.57 '.-.-..,. 1.878349 
 
 So is tang. ^^^_AM^ 5^0 ^y .... 10.192195 
 
 ,^ AEB ABE „ . r- -r.rnAn 
 
 To tang. ^ 18° 33' - - - - 9.o25943 
 
 ABE =38° 44' 
 
 In ABD-we have ABD=180°— 38° 44'=141° 16', ADB=13° 30', 
 and BAD=38° 44'— 13° 30'=25° 14'. To find AD and DB : 
 
 As sin. ADB 13° 30'. Ar. Co. 0.631815 
 
 Is to sin. BAD 25° 14' - - - 9.629721 
 
 So is AB 213 - - 2.328380 
 
 To BD 389- -.--.- 2.589916 
 
20 PLANE TRIGONOMETRY. 
 
 And, 
 
 As sin. ADB 13° 30' Ar. Co. 0.631815 
 
 Is to sin. ABD 141° 16' 9.796364 
 
 So is AB 213 - : 2.328380 
 
 To AD 570.9 - - - 2.756559 
 
 Finally, in ADC we have the angle ADC=43° 20', CAD=BAC + 
 HAD =60° 50' and the side AC ; to find CD. Thus, 
 
 As sin. ADC 43° 20' Ar. Co. 0.163523 
 
 Is to sin. CAD 60° 50' - 9.941117 
 
 So is AC 404 2.606381 
 
 To CD 514.1 2.711021 
 
 This might have been solved by finding ACB = 28° 14', CE = 
 292.87,. whence CBE would have been found to be = 77° 26', BD= 
 388.9, DC = 514, and AD = 570.8. 
 
 Example 5. (PI. 1, Fig. 9.) 
 
 Here AD = /BD^— AB' = v/ 1296 = 36. 
 And AC = AD + DC = 75 Ans. 
 
 Or, Trigbnometrically; 
 
 As BD .39 - - - Ar. Co. 8.408935 
 
 Is to BA 15 1.176091 
 
 So is radius '- lO.OOOOOO 
 
 To COS. B 67° 23' - 9.585026 
 
 And, 
 
 As radius ' . . . Ar. Co. 0.000000 
 
 Is to sin. B 67° 23' - 9.965248 
 
 So is BD 39 1.591065 
 
 To AD 36 1.556313 
 
 AC - 75 
 
PRACTICAL QUESTIONS. 21 
 
 Example 6. (PI. 1, fig. 10.) 
 
 The angle ACB = DBC — DAC = 25°. 
 Then, 
 
 As sin ACB 25° Ar. Co. 0.374052 
 
 * Is to sin. BAG 26° 30' 9.649527 
 
 So is AB 75 1.875061 
 
 To BC 79.18 1.898640 
 
 To find CD, and BD: 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. B 51° 30' - - 9.893544 
 
 So is CB - - - - - - - 1.898640 
 
 To CD 61.97 ^[^^ 
 
 And, 
 
 As radius Ar. Co. 0.000000 
 
 Is to COS. B ' 9.794150 
 
 So is CB 1.898640 
 
 To BD 49.29. - - - - - . - . . - . 1.692790 
 
 Example 7. (PI. 1, fig. 11.) 
 
 Here ACB = CAD = 35° and BAC = 55° 
 Hence, 
 
 As rad. Ar. Co. 0.000000 
 
 Is to tan. BAC 55° - - 10.154773 
 
 So is AB 143. 2.155336 
 
 To BC 204.2 2.310109 
 
 Example 8. (PI. 1, fig. 12.) 
 Construction. 
 
 Make AB=76, the distance from the lower column to the statue's 
 base. Erect the perpendiculars AD and BF, making the former= 
 50. With D as a centre and distance 86, cross BF in F, which 
 will be the head of the statue. 
 
22 PLANE TRIGONOMETRY. 
 
 Make AI = 64, draw IE parallel to AC, with F as a centre and 
 distance 97, cross IE in E, then EC perpendicular to AC, will be the 
 higher column. 
 
 Calculation. 
 
 To find FDG and side DG: 
 
 As DF 86 Ar. Co. 8.065502 
 
 Is to FG 76 1.880814 
 
 So is radius 10.000000 
 
 To sin. FDG 62° 5i' - 9.946316 
 
 As radius Ar. Co. 0.000000 
 
 Is to cos. FDG 62° oi' ------- - 9.670300 
 
 So is FD 86 1.934498 ' 
 
 To DG 40.25 1.604798 
 
 To find EFH and FH, we have FE= 97 and EH = GI = GD4 
 DI= 54.25. Hence, 
 
 As EF 97 Ar. Co. 8.013228 
 
 Is to EH 54.25 - 1.734400 
 
 So is radius - - - 10.000000 
 
 To sin. EFH 34° 9.747628 
 
 4nd, 
 
 iVs radius - Ar. Co. 0.000000 
 
 is to cos. F 34° • - 9.918574 
 
 So is EF 97 1.986772 
 
 To FH 80.42 1.905346 
 
 To find ED, we have EI = HF + FG= 156.42 and DI=U, 
 Hence, 
 
 As IE 156.42 Ar. Co. 7.805707 
 
 Is to ID 14 1.146128 
 
 So is radius - . 10.000000 
 
 To tan. lED 5° 7' - 8.951835 
 
 / 
 
PRACTICAL QUESTIONS. 23 
 
 As COS. E 5° 7' - Ar. Co. 0.001734 
 
 Is to rad. - 10.000000 
 
 So is IE 156.42 - - - - 2.194293 
 
 To ED 157.04 - - 2.196027 
 
 Otherwise. 
 
 GD = /FD^— FG» = ^/ 1620 = 40.25. 
 
 GI=GD + DI = 54.25. 
 
 FH = V'FE^— Eff = v/ 6465.9375 = 80.41. 
 
 IE = FH + FG = 156.41. 
 
 DPi = yiE^ + ID='= V 24660.0881 = 157.03. 
 
24 SURVEYING. [Cha!-. I. 
 
 SURVEYING. 
 
 CHAPTER I. 
 DIMENSIONS OF A SURVEY. 
 
 PROBLEM 8. 
 
 Example 2. 
 
 Angle B = 34° + 35° = 69°. 
 
 Example 3. 
 
 Here the first bearing must be reversed, since it is towards th t 
 station C. It becomes N. 35° W. Hence C= 180°— (35° +87°) =58^. 
 
 Example 4. 
 D = 180° ~ (87° — 58°) = 151°. 
 
 PROBLEM 9. 
 Example 2. 
 
 1st side S. 
 
 40J° 
 
 E. 
 
 3d 
 
 N. 29i° 
 
 E. 
 
 N 
 
 54 
 
 E. 
 
 
 N 54 
 
 E. 
 
 
 94i 
 
 
 
 N. 241 
 
 W. 
 
 
 180 
 
 
 
 
 
 N. 85i E. 
 
Prob. 8.] DIMENSIONSOFASURVEY. 25 
 
 
 4th 
 
 N. 281° E. 
 
 
 
 5th N. 57° W. 
 
 
 
 N. 54 E. 
 
 
 
 
 N. 54 E. 
 
 
 
 N. 25i W. 
 
 
 
 
 Ill 
 
 
 
 
 
 
 
 180 
 
 - 
 
 
 
 
 
 
 S. 69 W. 
 
 
 /' 
 
 6th 
 
 S. 
 
 470 
 
 W. 
 
 
 
 
 
 N. 54 
 
 E. 
 
 
 
 
 
 S. 
 
 7 
 
 E. 
 
 
 
 
 Example 
 
 3. 
 
 
 
 1st 
 
 S. 45i° W. 
 S. 20i W. 
 
 S. 25 W. 
 
 
 
 2d 
 
 N. 50° W. 
 
 S. 20i W. 
 
 N. 70i W. 
 
 
 3d 
 
 N. 0° W. 
 S. 20i W. 
 
 N. 20i W. 
 
 
 
 4th 
 
 N. 85° E. 
 S. 20J W. 
 
 N. 64i E. 
 
 
 5th 
 
 S. 47° E. 
 S. 20i W 
 
 S. 67i E. 
 
 
 
 7th 
 
 N. 51 i° W. 
 S. 20i W. 
 
 N. 711 W. 
 
 PROBLEM 10. 
 
 Example 1. 
 
 As sin. bearing 32° 30' - Ar. Co. 0.269784 
 
 Is to radius - ^ - 10.000000 
 
 So is departure 10.96 1.039811 
 
 To distance 20.40 1.309595 
 
 And, 
 
 As -radius - - - - Ar. Co. 0.000000 
 
 Is to cotangent bearing - 10.195813 
 
 So is departure 1.039811 
 
 To difference of latitude 17.20 - 1.235624 
 
20 SUKVEYING. [Chap. I. 
 
 Example 2. 
 
 As distance 44 Ar. Co. 8.356547 
 
 Is to difference of latitude 34.43 1.536937 
 
 So is radius 10.000000 
 
 To cosine of bearing 38° 31' 9.893484 
 
 And, 
 
 As rad. - - - Ar. Co. 0.000000 
 
 Is to tang, bearing 38° 31' 9.900864 
 
 So is diff. lat. 34.43 - 1.536937 
 
 To departure 27.40 1.437801 
 
 Example 3. 
 
 As cosine of bearing 32° 30' - - - - Ar. Co. 0.073971 
 
 Is to radius 10.000000 
 
 So is diff of lat. 17.21 1.235781 
 
 To distance 20.41 1.309752 
 
 And, 
 
 As radius Ar. Co. 0.000000 
 
 Is to tang, bearing 32° 30' 9.804187 
 
 So is diff latitude 17.21 .-..---- 1.235781 
 
 To departure 10.96 1.039968 
 
 Example 4. 
 
 As diff of lat. 27.92 N. ..... Ar. Co. 8.554085 
 
 Is to departure 5.32 E. - - - - 0.725912 
 
 So is radius 10.000000 
 
 To tang. bear. 10° 47' - - - 9.279997 
 
 And, 
 
 As cosine bearing 10° 47' Ar. Co. 0.007737 
 
 Is to radius 10.000000 
 
 So is diff of lat. 1.445915 
 
 To dist. 28.42 1.453652 
 
 / 
 
Prob. 12.] DIMENSIONS OF A SURVEY. 
 
 27 
 
 And, 
 
 Example 5. 
 
 As distance 35.35 Ar. Co. 8.451611 
 
 Is to departure 15.08 - - 1.178401 
 
 So is radius - - 10.000000 
 
 To sin. bearing 25° 15' ....... 9.630012 
 
 As radius .-.------ Ar. Co. 0.000000 
 
 Is to cos. bearing 25° 15' 9.956387 
 
 So is distance 1.548389 
 
 To diff. of lat. 31.97 1.504776 
 
 PROBLEM 12. 
 
 Sta. 
 
 Courses. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 Cor. 
 
 N. 
 
 Cor. 
 E. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 1 
 
 N. 75 E. 
 
 13.70 
 
 3.54 
 
 
 13.24 
 
 
 2 
 
 2 
 
 3.56 
 
 
 13.26 
 
 
 2 
 
 N. 20| E. 
 
 10.30 
 
 9.65 
 
 
 3.61 
 
 
 1 
 
 1 
 
 9.66 
 
 
 3.62 
 
 
 3 
 
 East. 
 
 16.20 
 
 
 
 16.20 
 
 
 2 
 
 2 
 
 .02 
 
 
 16.22 
 
 
 4 
 
 S. 33i W. 
 
 35.30 
 
 
 29.44 
 
 
 19.49 
 
 5 
 
 5 
 
 
 29.39 
 
 
 19.44 
 
 5 
 
 S. 76 W. 
 
 16.00 
 
 
 3.87 
 
 
 15.52 
 
 2 
 
 2 
 
 
 3.85 
 
 
 15.50 
 
 6 
 
 North. 
 
 9.00 
 
 9.00 
 
 
 
 
 1 
 
 1 
 
 9.01 
 
 
 .01 
 
 
 7 
 
 S. 84W. 
 
 11.60 
 
 
 1.21 
 
 
 11.54 
 
 2 
 
 2 
 
 
 1.19 
 
 
 11.52 
 
 8 
 
 N. 531 w. 
 
 11.60 
 
 6.94 
 
 
 
 9.29 
 
 2 
 
 2 
 
 6.96 
 
 
 
 9.27 
 
 9 
 
 N. 363 E. 
 
 19.36 
 
 15.51 
 
 
 11.59 
 
 
 3 
 
 2 
 
 15.54 
 
 
 11.61 
 
 
 10 
 
 N. 224 E. 
 
 14.00 
 
 12.93 
 
 
 5.36 
 
 
 2 
 
 2 
 
 12.95 
 
 
 5.3S 
 
 
 11 
 
 S. 76| E. 
 
 12.00 
 
 
 2.75 
 
 11.68 
 
 
 2 
 
 2 
 
 
 2.73 
 
 11.70 
 
 
 12 
 
 S. 15 W. 
 
 10.85 
 
 
 10.48 
 
 
 2.81 
 
 2 
 
 1 
 
 
 10.46 
 
 
 2.80 
 
 13 
 
 S. 18 W. 
 
 10.62 
 
 
 10.10 
 
 
 3.28 
 
 2 
 
 1 
 
 
 10.08 
 
 
 3.27 
 
 
 
 
 57.57 
 
 57.85 
 57.57 
 
 61.68 
 
 61.93 
 61.68 
 
 
 
 
 
 
 
 Error South .28 
 
 .25 Error West- 
 
28 
 
 SURVEYING. 
 
 [Chap. II. 
 
 CHAPTER 11. 
 SUPPLYING OMISSIONS. 
 
 . PROBLEM I. 
 Example 2. 
 
 Sta. 
 
 Courses. 
 
 Dist. 
 
 N. 
 
 s. 
 
 E. 
 
 W. 
 
 1 
 
 N. i5r w. 
 
 9.40 
 
 9.05 
 
 
 
 2.55 
 
 2 
 
 N. 631 E. 
 
 10,43 
 
 4.61 
 
 
 9.36 
 
 
 3 
 
 S. 49 E. 
 
 8.12 
 
 
 5.33 
 
 6.13 
 
 
 4 
 
 S. 13i E. 
 
 8.45 
 
 
 8.22 
 
 1.98 
 
 
 5 
 
 S. 161 E. 
 
 6.44 
 
 
 6.17 
 
 1.86 
 
 
 6 
 
 
 
 
 (6.11) 
 
 
 (10.64) 
 
 7 
 
 N. 60 W. 
 
 9.72 
 
 4.86 
 
 
 
 8.41 
 
 8 
 
 N. 17i W. 
 
 7.65 
 
 7.31 
 
 
 2.27 
 
 
 
 
 
 25.83 
 
 25.83 
 
 21.60 
 
 21.60 
 
 Then, 
 
 As diff. lat. 6.11 S. Ar. Co. 9.213959 
 
 Is to depart. 10.64 W. - - 1.026942 
 
 So is radius 10.000000 
 
 To tang, bearing S. 60° 8' W. 
 
 10.240901 
 
 And, As cosine bearing 60° 8' Ar. Co. 0.302785 
 
 Is to radius - 10.000000 
 
 So is diff. lat. 0.786041 
 
 To distance 12.27 1.088826 
 
 Example 3. 
 
 Sta. 
 
 Courses. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 1 
 
 S. 52° W. 
 
 10.70 
 
 
 6.59 
 
 
 8.43 
 
 2 
 
 S. 7i W. 
 
 13.92 
 
 
 13.80 
 
 
 1.82 
 
 3 
 
 S. 34i E. 
 
 9.00 
 
 
 7.44 
 
 5.07 
 
 
 4 
 
 
 
 (27.83) 
 
 
 (5.18) 
 
 
 
 
 
 
 27.83 
 
 10.25 
 
 10.25 
 
Prob. 1.] 
 
 SUPPLYING OMISSIONS. 
 
 29 
 
 Then, 
 
 As diff. lat. 27.83 Ar. Co. 8.555487 
 
 Is to departure 5.18 ----- 0.714330 
 
 So is radi-us 0.000000 
 
 To tang, bearing N. 10° 33' E. 9.269817 
 
 And, 
 
 As cosine bearing 10° 33' - - - - - Ar. Co. 0.007404 
 
 Is to radius 10.000000 
 
 So is diff. lat. 27.83 1.444513 
 
 To distance 28.31 1.451917 
 
 Example 4. 
 
 Sta. 
 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 1 
 
 S. 10° E. 
 
 92.20 
 
 
 90.80 
 
 16.01 
 
 
 2 
 
 S. 15 W. 
 
 120.50 
 
 
 116.39 
 
 
 31.19 
 
 3 
 
 s. m w. 
 
 205.00 
 
 
 194.40 
 
 
 65.05 
 
 4 
 
 8. 7H E. 
 
 68.00 
 
 
 21.58 
 
 64.49 
 
 
 5 
 
 
 
 
 
 
 
 
 
 
 
 423.17 
 
 80.50 
 
 96.24 
 80.50 
 
 15.74 
 
 Then, 
 
 As diff. of latitude 423.17 
 Is to departure.,15.74 - - 
 So is radius ----- 
 
 Ar. Co. 7.373485 
 
 - - - 1.197005 
 
 - - - 10.000000 
 
 To tans;, bearina; 2° 8' 
 
 8.570490 
 
 And, 
 
 As cosine bearing 2° 8' 
 Is to radius - - 'i - 
 So is diff lat. 423.17 - 
 
 Ar. Co. 0.000301 
 
 - - - 10.000000 
 
 - - - 2.626515 
 
 To distance 423.46 
 
 2.626816 
 
30 
 
 SURVEYING. 
 
 [Chap. II. 
 
 PROBLEM 11. 
 Example 2. 
 
 Sta. 
 
 Bearing. 
 
 Changed 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. W. 
 
 1 
 
 S. 40i E. 
 
 N. 85i E." 
 
 31.80 
 
 2.49 
 
 
 31.70 
 
 
 2 
 
 N. 54 E. 
 
 North. 
 
 
 (2.08) 
 
 
 
 
 3 
 
 N. 29i E. 
 
 N. 24f W. 
 
 2.21 
 
 2.01 
 
 
 
 .93 
 
 4 
 
 N. 28| E. 
 
 N. 25i W. 
 
 35.35 
 
 31.98 
 
 
 
 15.08 
 
 5 
 
 N. 57 W. 
 
 S. 69 W. 
 
 
 
 (7.49) 
 
 
 (1^.51) 
 
 6 
 
 S. 47 W. 
 
 S. 7 E. 
 
 31.30 
 
 
 31.07 
 
 3.82 
 
 
 
 
 
 
 38.56 
 
 38.56 
 
 35.52 
 
 35.52 
 
 As sine changed bearing 69° 
 Is to radius ------ 
 
 So is departure 19.51 - - - 
 
 To distance 5th side 20.90 - 
 
 Ar.Co. 0.029848 
 
 - - - 10.000000 
 
 - - - 1.290257 
 
 1.320105 
 
 A.nd, As radius 
 
 Is to cotaug. bearing 69° 
 So is departure 19.51 - - 
 
 To diff. latitude 7.49 S. - 
 
 Ar. Co. 0.000000 
 
 - - - 9.584177 
 
 - - - 1.290257 
 
 0.874434 
 
 PROBLEM III. 
 Example 2. 
 
 6ta. 
 
 Bearing. 
 
 Changed 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 . E. 
 
 W. 
 
 1 
 
 S. 40i E. 
 
 N. 85i E. 
 
 31.80 
 
 2.49 
 
 
 31.70 
 
 
 2 
 
 N. 54 E. 
 
 North. 
 
 
 (2.09) 
 
 
 
 
 3 
 
 N. 29i E. 
 
 N.24|W. 
 
 2.21 
 
 2.01 
 
 
 
 .93 
 
 4 
 
 N. E. 
 
 
 35.35 
 
 (31.97) 
 
 
 
 (15.08) 
 
 5 
 
 N. 57 W. 
 
 S. 69 W. 
 
 20.90 
 
 
 7.49 
 
 
 19.51 
 
 6 
 
 S. 47 W. 
 
 S. 7 E. 
 
 31.30 
 
 
 31.07 
 
 3.82 
 
 
 
 
 
 
 38.56 
 
 38.56 
 
 35.52 
 
 35.52 
 
Prob. 4.] 
 
 SUPPLYING OMISSIONS. 
 
 31 
 
 Then, 
 
 As distance 4th side 35.35 - - - Ar. Co. 8.451611 
 
 Is to departure 15.08 1.178401 
 
 So is radius 10.000000 
 
 To sine chang. bearing N. 25° 15' W. 
 
 54 
 
 9.630012 
 
 Bearing 4th side 
 
 N. 28° 45' E. 
 
 And, 
 
 As radius - - - - Ar. Co. 0.000000 
 
 Is to cos. chang. bearing 25° 15' .... 9.956387 
 So is distance ---.-- 1.548389 
 
 To diff. latitude 31.97 1.504776 
 
 PROBLEM IV. 
 
 Example 2. (PI. 1, fig. 13.) 
 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 FA 
 
 S. E. 
 
 31.80 
 
 
 
 
 
 AB 
 
 N. 54 E. 
 
 2.08 
 
 1.23 
 
 
 1.68 
 
 
 BC 
 
 N. 29i E. 
 
 2.21 
 
 1.92 
 
 
 1.08 
 
 
 CD 
 
 N. 281 E. 
 
 35.35 
 
 31.00 
 
 
 17.00 
 
 
 DE 
 
 N. 57 W. 
 
 20.90 
 
 11.38 
 
 
 
 17.52 
 
 EF 
 
 S. W. 
 
 31.30 
 
 
 
 
 
 Diff. 
 
 latitude of '. 
 
 EA 
 
 45.53 
 
 
 19.76 
 17.52 
 
 17.52 
 
 Departure of EA 2.24 
 Then, As diff. lat. EA 45.53 . - . . Ar. Co. 8.341702 
 
 Is to departure 2.^4 0.350248 
 
 So is radius 10.000000 
 
 To tang, bearing EA 2° 49' 
 
 8.691950 
 
32 SURVEYING. [Chap. II. 
 
 And, 
 
 As cosine bearing 2° 49' ----- Ar. Co. 0.000525 
 
 Is to radius - 10.000000 
 
 So is diff. lat. - - . - - 1.658298 
 
 To distance EA 45.59 - - - - 1.658823 
 
 To find AEF: 
 
 AF 31.80 
 
 AE 45.59 Ar. Co. 8.341177 
 
 EF 31.30 Ar. Co. 8.504456 
 2)108.69 
 
 Half sum 54.34 1.735120 
 
 Difference 22.54 ------ 1.352954 
 
 """^" 2 )19.933707 
 
 Cos. i AEF 22° 6' - - - 9.966853 
 
 AEF 44° 12' 
 
 Bearing of EA - 2° 49' 
 
 EF S.47°~TW. 
 
 To find EAF and bearing of FA : 
 
 As AF 31.80 Ar. Co. 8.497573 
 
 Is to EF 31.30 1.495544 
 
 So is sin. AEF 44° 12' 9.843336 
 
 To sin. EAF - 43° 20' 9.836453 
 
 Bearing of EA 2° 49' ==" 
 
 AF 40° 31' 
 
 CHAPTER III. 
 CONTENT OF LAND. 
 
 PROBLEM I. 
 
 Example 4. 
 
 Here, Area = 176.4 x 176.4 = 31 1 16.96 Sq. Perches, 
 
 = 194 A. 1 R. 36^96 P. 
 
Proe. 3.] CONTENT OF LAND. 38 
 
 Example 5. 
 
 Here, Area = 52.25 x 38.24 = 1998.04 Sq. Ch, 
 = 199 A. 3 R. 8.64 P. 
 
 Example 6. 
 
 Here, Area = 16.54 x 12.37 = 204.5998 Sq. Ck 
 = 20 A. 1 R. 33.5968 P. 
 
 Example 7. 
 
 Here, Area = 21.16 x 11.32=239.5312 Sq. Ch. 
 = 23 A. 3 R. 32.4992 P. 
 
 PROBLEM 2. 
 
 Example 2 
 
 . 18.37X13.44 246.8928 ,^^..^, ^ ^, 
 ;re, Area = = = 123.4464 Sq. Ch. 
 
 = 12 A. 1 R. 15.1424 P. 
 
 Example 3. 
 
 49X34 1666 ^ „ 
 
 Here, Area = — ^ — = ~2~ "^ ^^^ ^' 
 
 = 5 A. R. 33 Pe. 
 
 PROBLEM 3. 
 
 •Example 2. (PI. 1, fig. L) 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. A 47° 30' 9.867631 
 
 AB 15.36 1.186391 
 
 AC 11.46 1.059185 
 
 So is AB X AC 
 
 To double area 129.78 2.113207 
 
 ABC - 64.89 Ch. = 6 A. 1 R. 38. 24 P 
 
 3 
 
34 SURVEYING. [Chap. Ill, 
 
 Example 3. (PL 1, fig. 14.) 
 
 Here, As radius - Ai\ Co. 0.000000 
 
 Is to sin. A 66° 30' 9.962398 
 
 ^ . ,^ ,„(AB 13.84 1.141136 
 
 boisABXAUj^^ 18.23 - 1.260787 
 
 To 2 ABC 231.38 - 2.364321 
 
 ABC - 115.69 Ch. = 11 A. 2 R. 1L04 R 
 
 Example 4. (PI. 1, fig. 15.) 
 
 Here, As radius .--.-..-. Ar. Co. 0.000000 
 
 Is to sin. A 121° 45' ------- - 9.929599 
 
 ^ ■ AH &n iAB 19.74 - - - - - 1.295347 
 
 feo IS AJJ, AO j ^^ ^^^^^ ..... 1.239049 
 
 To 2 ABC 291,07 2.463995 
 
 ABC - 145.535 Ch. = 14 A. 2 R. 8.56 P. ^^"""^ 
 
 PROBLEM 4. 
 
 Example 2. (PL 1, fig. 1.) 
 Here, Angle C = 180— (A + B) = 43°. Hence, 
 
 A A • n \ radius - - - Ar. Co. 0.000000 
 
 As rad., sm. C | ^.^^ ^ ^^, _ ^^^ ^^^ ^^^^^^^^ 
 
 , , . A • Tj < sin. A 63° - - - . 9.949881 
 
 Is to sm. A, sm. B j ^.^^ ^ ^^, _ _ ^^^^^^^^ 
 
 ^ . ,^2 (AB 24.32 - - - - 1.385964 
 
 feo is At. i AB ------ - 1.385964 
 
 To 2 ABC 742.8 2.870868 
 
 ABC - ^L4 Ch. r= 37 A. R. 22.4 P. • """""^ 
 
 Example 3. 
 Here, the angle C = 94° 15'. Hence 
 
 A A • r i rad. - - - Ar. Co. 0.000000 
 
 As rad., sm. C j ^.^^ ^ ^^, ^^, ^^^ ^^^ ^^^^^^^^ 
 
 T , . A • -D < sin. A 37° 30' - - - 9.784447 
 
 Is to sm. A. sm. B < . „ „ ^ onL^nrm 
 
 I sin. B 48° 15' - - - 9.872772 
 
 ^ • ,r,, (AB 17.36 - - - - 1.239550 
 
 f^o IS AtS j ^g 1.239550 
 
 To 2 ABC 137.25 -...-... 2.137515 
 ABC -~68!625 Ch. = 6 A. 3 R. 18 P. 
 
 / 
 
Prob. 6.] CONTENT OF LAND. 35 
 
 PROBLEM 5. 
 
 Example 2. 
 Here, 10.64 + 12.28 + 9.00 = 3L92 = sum of sides. 
 
 Half sum 15.96 log. 1.203033 
 
 r 5.32 0.725912 
 
 Remainders ) 3.68 0.565848 
 
 ( 6.96 0.842609 
 
 2)8.837402 
 
 Area 10)46.63 Ch. ..----- 1.668701 
 
 4.663 = 4 A. 2 R. 26.08 P. 
 
 Example 3. 
 
 Here, 20 + 30 + 40 = 90 = sum of sides. ' 
 Half sum 45 - - 1.653218 
 
 /25 1.397940 
 
 Remainders <15 1.176091 
 
 (5 0.698970 
 
 2)4.926214 
 
 10)290.47" 2.463107 
 
 29,047 A. = 29 A. R. 7.52 P. 
 
 PROBLEM 6. 
 
 Example 2, 
 
 Here, 16.10 x ^4^^ = 16.1x5.1 = 82.11 Ch. 
 
 = 8 A. OR. 33.76 P. 
 
 Example 3. 
 
 8 274- 12 43 
 Here, 24 X - — ^ — '— = 24 X 10.35 = 248.4 Ch. 
 
 = 24 A. 3R. 14.4 P. 
 
36 SURVEYING. [Chap. III. 
 
 PROBLEM 7. 
 Example 2. 
 
 12.41+8.22 
 Here, Area = ~ x 5.15 = 53.12225 Ch. 
 
 = 5 A. 1 R. 9.95G P. 
 
 Example 3. 
 
 11.34+18.46 
 Here, Area = ~ x 13.25 = 197.425 Ch. 
 
 = 19 A. 2R. 38.8 P 
 
Prob. 9.] 
 
 CONTENT OF LAND. 
 
 37 
 
 pq 
 O 
 
 
 C3 
 
 
 
 00 
 
 CO 
 
 »o 
 It 
 
 T— 1 
 
 1 
 
 1—; 
 
 
 CO 
 
 CO 
 
 tHO 
 
 (:oco 
 
 CO CO 
 CO t^ 
 
 CO 00 
 
 r-l 
 
 U 
 
 o 
 
 CO 
 CO 
 lO 
 
 T— 1 
 
 o 
 o 
 
 CO 
 
 
 
 o 
 o 
 
 o 
 co' 
 
 
 o 
 
 CO 
 CO 
 
 oo' 
 
 ft 
 
 ft 
 
 o 
 c4 
 
 o 
 o 
 
 r-l 
 
 o 
 "*. 
 
 co' 
 1—1 
 
 00 
 
 o 
 
 1—1 
 
 05 
 
 
 
 ^ 
 
 o 
 
 CO 
 
 
 
 00 
 
 o 
 I— 1 
 
 OS 
 
 CO 
 
 CO 
 
 o 
 
 T— I 
 
 H 
 
 
 O 
 
 o 
 
 CO 
 
 
 
 
 o 
 1—1 
 
 02 
 
 
 
 
 CO 
 
 
 ^ 
 
 rH 
 
 o 
 
 1—1 
 
 ;zi 
 
 1— ( 
 Oi 
 
 OQ 
 l-H 
 
 
 
 o 
 c<i 
 
 
 T— 1 
 
 o 
 1—1 
 
 u 
 o 
 
 o 
 
 o 
 
 rH 
 
 o 
 
 o 
 
 o 
 
 T— I 
 
 a5 
 
 6 
 
 o 
 
 o 
 
 T-i 
 
 o 
 
 o 
 
 o 
 
 I— 1 
 
 ^ 
 
 o 
 
 CO 
 
 
 
 00 
 
 o 
 
 1—5 
 
 CO 
 CO 
 
 o 
 I— 1 
 
 w 
 
 
 o 
 o 
 
 CO 
 
 
 
 
 o cr> 
 1—1 1— 1 
 
 03 
 
 
 
 1—1 
 
 CO 
 
 1-H 
 
 
 05 
 
 o 
 1—1 
 
 ^ 
 
 1—1 
 
 00 
 
 I-I 
 
 
 
 o 
 o4 
 
 
 1—1 1—1 
 
 d d 
 1—1 1— ( 
 
 "S 
 
 CO 
 
 
 oo 
 
 CO 
 
 Oi 
 CO 
 
 CO 
 
 co" 
 
 o 
 o 
 d 
 CO 
 
 
 V 
 
 o 
 
 00 
 
 CO 
 
 r-l 
 
 o 
 
 IC 
 
 CO 
 
 W 
 
 V 
 05 
 
 •o 
 
 >— 1 
 
 CO 
 
 o 
 
 CO 
 CO 
 
 V 
 
 CO 
 
 V 
 05 
 lO 
 o 
 
 CO 
 CO 
 
 
 
 ■ 1— 1 
 
 C<) 
 
 eo 
 
 'tt 
 
 lO 
 
 CO 
 
 
 1—1 
 
 o 
 
 
 
 o 
 
 o 
 
 
 o 
 
 o 
 
 o 
 
 ICTtl 
 
 o^ 
 
 Ci 
 
 ■^ 
 
 ■*! 
 
 oo 
 
 TtH 
 
 t^ 
 
 t- 
 
 o 
 
 2. 
 
 -^ 
 
 -i* 
 
 c-i 
 
 Ph 
 
 CO 
 
 of 
 
 w 
 
 OQ 
 
 W 
 
38 
 
 
 
 
 
 
 
 
 
 
 SURVEYING. 
 
 
 
 
 
 
 [Chap. III. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 o 
 
 o 
 
 
 
 Tt< 
 
 
 
 
 
 
 
 CO 
 
 m 
 
 
 
 
 CO 
 
 Oi 
 
 O 
 
 7-< 
 
 
 TO t^ 
 
 CO 
 
 TO 
 
 TO tH CMTf< 
 
 00 
 
 
 03 
 
 OS 
 
 
 
 
 
 
 
 CO 
 
 rH 
 
 
 
 
 CO 
 
 lO 
 
 OO 
 
 TO 
 
 
 rHO 
 
 o 
 
 in 
 
 lO T—< 
 
 TfH 
 
 
 
 1—1 
 
 
 
 
 
 
 
 o 
 
 ^ 
 
 
 
 
 TO 
 
 (M 
 
 00 
 
 as 
 
 
 rH (M 
 
 as 
 
 ■^ 
 
 ^ 00 
 
 C<J 
 
 
 T-J 
 
 
 
 
 
 
 
 00 
 
 -^ 
 
 
 
 
 CO 
 
 CO 
 
 CO 
 
 00 
 
 
 <^. ^. 
 
 CO 
 
 CO 
 
 TO TO 
 
 IC 
 
 
 < 
 
 t>I 
 
 
 
 
 
 
 
 id 
 
 '^' 
 
 
 
 
 rH* 
 
 OO' 
 
 id 
 
 ^ 
 
 
 CO id 
 
 C4 
 
 CD 
 
 CO in 
 
 rH 
 
 
 
 00 
 
 
 
 
 
 
 
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Pros. 9.] 
 
 CONTENT OF LAND. 
 
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 As diff. lat 11.24 Ar. Co. 8.949234 
 
 Is to departure 25.71 1.410102 
 
 So is radius ^ 10.000000 
 
 To tang, bearing 66° 23' - 10.359336 
 
 And, 
 
 Ab cosine bearing 66° 23' Ar. Co. 0.397272 
 
 Is to radius - - - 10.000000 
 
 So is diff. lat. 11.24 1.050766 
 
 To distance 28.06 1.448038 
 
40 
 
 SURVEYING. 
 
 [Chap. IIL 
 
 
 
 OQ 
 
 
 W i 
 
 W 
 
 o 
 
 -IN 
 
 I— I 
 
 o b- 
 CO 
 
 
 >o 
 
 lOrtH 
 
 oo 
 
 rH 
 
 o 
 
 
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Prob. 9.j C N T E N T F L A N D . 41 
 
 As difF. lat. DF 3.91 Ar. Co. 9.407823 
 
 Is to depart. 19.73 1.295127 
 
 So is radius .....-- 0.000000 
 
 To tang, bearing N. 78° 47' E. - - - - - 10.702950 
 
 As cosine bearing 78° 47' - - - Ar. Co. 0.711036 
 
 Is to radius - 10.000000 
 
 So is diff. latitude 3.91 0.592177 
 
 To distance DF 20.10 r - 1.303213 
 
 FE 19.18 
 
 ED 9.61 --"...■- Ar. Co. 9.017277 
 
 DF 20.10 .-.---." " 8.696804 
 
 2)48.89 
 
 Half sum 24.445 - - - - 1.388190 
 
 Diff. 5.265 ----.--•-- 0.721398 
 
 2)19.823669 
 
 Cos. i FDE 35° 17' 9.911834 
 
 FDE 70° 34' 
 
 Bearing DF N. 78 47 E. 
 
 Bearing DE N. 8° 13' E. 
 
 As FE 19.18 - - -. -8.717151 
 
 Is to DE 9.61 --...-.... 0.982723 
 
 So is sin. FDE 70° 34' -.-.... 9.974525 
 
 To sin. DFE" - - 28° 12' 9.674399 
 
 Bearing FD S. 78 47 W. 
 
 106 59 
 180 
 
 Bearing EF S. 73 1 E. 
 
42 
 
 SURVEYING. 
 
 [Chap. III. 
 
 Example 6. (Fig. 81, Surveying.') 
 
 
 
 
 
 
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 00 
 
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 T-H 
 
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 00 
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 00 
 
 c<i 
 
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 II 
 
 1 
 
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 00 
 
 p4 
 
 1—1 
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 p4 
 
 rH 
 
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 00 
 
 p4 
 
 Ml* 
 
 OO 
 
 CO 
 
 Ml* 
 
 o 
 
 C<J 
 
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 rH 
 
 
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 rtkM 
 
 o 
 
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 o 
 
 e4 
 
 00 
 
 
 03 
 
 f4 
 
 o 
 
 CO 
 
 ■m 
 
 p4 
 
 CO 
 
 o 
 m 
 
 
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 p 
 
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 M 
 1— 1 
 
 
 t 
 
 
Prob. 9.] CONTENT OF LAND. 
 
 As sine changed bearing LA 71° 45' - - Ar. Co. 0.022414 
 
 Is to radius 10.000000 
 
 So is departure 22.61 1.354301 
 
 To distance LA 23.81 1.876715 
 
 And, 
 
 As radius Ar. Co. 0.000000 
 
 Is to cotang. bearing 9.518184 
 
 So is departure - 1.354301 
 
 To diff. latitude 7.46 - - 0.872485 
 
 43 
 
 Example 7. (Fig. SOj Surveying.) 
 To find the third side. 
 
 Sta. 
 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 EA 
 
 S. 52 W. 
 
 10.70 
 
 
 6.59 
 
 
 8.43 
 
 AB 
 
 S. 7i W. 
 
 13.92 
 
 
 13.80 
 
 
 1.82 
 
 BC 
 
 S. 33i E. 
 
 9.00 
 
 
 7.53 
 
 4.93 
 
 
 
 
 
 
 27.92 
 
 
 10.25 
 4.93 
 
 5.32 
 
 As diff. lat. EC 27.92 
 Is to depart. 5.32 - 
 So is radius - - - 
 
 Ar.Co. 8.554085 
 
 - - - 0.725912 
 
 - - - 10.000000 
 
 To tang, bearing S. 10° 47' W. 
 
 As cosine bearing 10° 47' - ■ 
 
 Is to radius ■ 
 
 So- is diff. lat. 
 
 9.279997 
 
 Ar.Co. 0.007737 
 
 - - - 10.000000 
 
 - - - 1.445915 
 
 To distance 28.42 - 1.453652 
 
44 
 
 SURVEYING. 
 
 [Chap. III. 
 
 I 1 
 I in ■* 
 
 o 
 
 IC ^ 
 
 o 
 
 lO 
 
 t- 
 
 oo 
 
 
 
 Oi -^ 
 
 Oi 
 
 ^ 
 
 Ci 
 
 (M 
 
 1—1 
 
 lO 
 
 o 
 
 o 
 
 
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 00 
 
 
 T-H 
 
 
 
 
 
 
 M 
 
 
 
 Pk 
 
 Ph 
 o 
 
 o 
 
 ^ CO 
 
 I I 
 
 o 
 
 CO 
 
 
 M 
 
 CD 
 
 ^ 
 
 ^ 
 S' 
 
 cq 
 
Prob. 10.] 
 
 CONTENT OF LAND. 
 
 45 
 
 PROBLEM 10. 
 
 Example 2. 
 
 
 
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 CD 
 
 § 
 
 o 
 
 
 
 0^ 
 
 
 
 
 
 
 
 o 
 
 in 00 
 
 f^ 
 
 00 t> 
 
 10 
 
 lOTl* 
 
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 00 o 
 
 03 
 
 
 
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 r-l 
 
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 o> 
 
 
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 ^ 
 
 w^ 
 
 l» 
 
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 fr: 
 
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 O 
 
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 W-f 
 
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 CO 
 
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 lO 
 
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 lO 
 
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 cr. 
 
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 rt -* 
 
 
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 rH 
 
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 w 
 
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 r^ 
 
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46 
 
 SUKVEYING. 
 
 [Chap. III. 
 
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PROB. li.J CONTENT OF LAND. 47 
 
 PROBLEM 13. 
 
 Example 2. (PI. 2, fig. 5.) 
 
 Here the various angles will be found to be as in the following 
 proportions. Then, 
 
 To find log. of GA : 
 
 As sin. FAG 88° 30' - - . . . Ar. Co. 0.000149 
 
 Is to sin. GFA 68° 30' 9.968678 
 
 So is FG 20 ch. - - 1.301030 
 
 To GA - 1.269857 
 
 To find log. GB: 
 
 As sin. FBG 42° x\r. Co. 0.174489 
 
 Is to sin. GFB 24° 9.609318 
 
 So is FG -1.301030 
 
 To GB 1.084832 
 
 To find log. GC: 
 
 As sin. GCF 43° 15' Ar. Co. 0.164193 
 
 Is to sin. GFC 38° 9.789342 
 
 So is FG 1.301030 
 
 To GC 1.254565 
 
 To find log. GD: 
 
 As sin. GDF 44° 30' ..... Ar. Co. 0.154338 
 
 Is to sin. GFD 59° - . ...... 9.933066 
 
 So is GF 1.301030 
 
 To GD - -^- - - - - - 1.388434 
 
 To find log. GE : 
 
 Assin. GEF 35° 30' ..... Ar. Co. 0.236046 
 
 Is.to sin. GFE 103° 30' 9.987832 
 
 So is GF 1.301030 
 
 To GE 1.524908 
 
48 SUKVEYING. [Chap. III. 
 
 To find 2ABG: 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. AGB 91° 9.999934 
 
 o • -or- Ai- ^BG 1.084832 
 
 So IS BG, AG j ^^ ^^^3^3^^ 
 
 To 2 ABG 226.268 2.354623 
 
 Tofind2BGC: 
 
 As radius Ar Co. 0.000000 
 
 Is to sin. BGC 15° 15' 9.420007 
 
 ^ ' rn rr i ^B - 1.084832 
 
 bo IS GB, GC I ^^ _ 1.254565 
 
 To 2 BGC 57.465 1-759404 . 
 
 To find 2CGD: 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin. CGD 22° 15' 9.578236 
 
 a • nn nn \ ^C 1.254565 
 
 So IS GC, GD I ^j^ 1.388434 
 
 To 2 CGD 166.431 2.221285 
 
 To find 2 DGE : 
 
 As radius ..-..-.. Ar. Co. 0.000000 
 
 Is to sin. DGE 35° 30' 9.763954 
 
 GD 1.388434 
 
 So is GD, GE ^ ^^ 1.524908 
 
 To 2 DGE 475.657 ----.--- 2.677296 
 
 To find 2 EGA : 
 
 As radius - - Ar. Co. 0.000000 
 
 Is to sin. EGA 18° 9.489982 
 
 ^n ^.vr CA ^GE 1.524908 
 
 So IS EG, GA I ^^ ^269857 
 
 To 2EGA 192.640 - 2.284747 
 
 2 DGE 475.657 [ — 
 
 2 CGD 166.431 
 
 2 BGC 57.465 
 
 892.193 
 2 AGB 226.268 
 
 2)665.925 
 
 ABCDE 332.9625 Ch. = 33 A. 1 R. 7.4 P. 
 
 / 
 
Prob. 5.] LAYING OUT AND DIVIDING LAND. 49 
 
 CHAPTER IV. 
 LAYING OUT AND DIVIDING LAND. 
 
 PROBLEM 1. 
 
 Example 2. 
 
 Here, 325 Acres = 3250 chains. 
 And side = v 3250 = 57 chains. 
 
 PROBLEM 2. 
 Example 2. 
 
 Here breadth = ^— ,- = — = 6.25 chains. 
 
 8 chains 8 
 
 PROBLEM 3. 
 
 Example 2. 
 
 Here, 27 A. 3 R. 20 P. = 4460 P. 
 
 And, As 7 : 9 : : 4460 : 5734.2857. 
 
 /5734.2857"== 75.725 = length. 
 
 Also, As 9 : 7 : : 75.725 : 58.897 = breadth. 
 
 PROBLEM 4, 
 
 Example 2. (PI. 2, fig. 6.) 
 
 Here, 114 A. 2 R. 33.4 P. = 1147.0875 chains. 
 Also, 71147.0875 + 7.55^ = v/ 1204.09 = 34.7. 
 And, 34.7 + 7.55 = 42.25 length. 
 34.7—7.55 = 27.15 breadth. 
 
 ^ PROBLEM 5. 
 Example 3. (PI. 2, fig. 7.) 
 Here, 2 Acres = 320 Perches. 
 
50 SURVEYING. [Chap. IV. 
 
 And, 
 
 . . P . . ( AB 30 P. - Ar. Co. 8.522879 
 
 AS At5, Sin, A j ^.^^ ^ ^j, jg, ^^^ ^^^ 0.023682 
 
 Is to 2 ABC 640 2.806180 
 
 So is radius - - 10.000000 
 
 To AC 22.53 1.352741 
 
 Example 4. (PI. 2, fig. 8.) 
 
 AB 32.26 - Ar. Co. 8.491336 
 
 As AB, sin. A . ^.^^ ^ ^^^ ^^, ^^ ^^_ 0.002801 
 
 Is to ABCD 740 2.869232 
 
 So is radius 10-000000 
 
 To AD 23.09 1.363369 
 
 PROBLEM 6. 
 
 Example 2. (PI. 2, fig. 9.) 
 
 Here, 27 A. IR. 16 P. = 273.5 Ch. 
 And, 
 
 As ABC 273.5 - Ar. Co. 7.563043 
 
 Is to BDC 100 2.000000 
 
 So is AB 35.20 1.546543 
 
 To BD 12.87 r - - - 1.109586 
 
 PROBLEM 7. 
 
 Example 2. (PI. 2, fig. 10.) 
 
 Construction. 
 
 Make AB, equal to the greater of the given sides (20). Draw 
 BD perpendicular to AB, equal to twice the given area, divided by 
 AB (12.39). Through D draw DC parallel to AB. Then if AC 
 be made equal to the other given side (16.25), and BC be joined; 
 ABC will be the triangle. 
 
 For the Division Line. Make AP = 8.50 the given distance. 
 Take AF to AC in the ratio of the part to be cut off to the whole 
 area. Join PF, draw BG parallel to it ; then PG will be the division 
 line. 
 
Pros. 9.] LAYING OUT AND DIVIDING LAND. 51 
 
 Demonstration. 
 AB : AP : : AG : AF, Therefore, AB. AC : AP. AG : : AC . AG : 
 AG.AF : : AC : AF, or AB.AC.sin. A : AP. AG.sia A : : AC : 
 AF : : m : n (m being the whole area, and n the part to be cut off.) 
 Hence, since AC . AB sin. A = m, AP. AG sin. A = n, and PG is 
 .the division line. 
 
 Calculation. 
 
 As ABC 123.9375 Ar. Co. 7.906798 
 
 Is to APG 30 1.477121 
 
 ^ ■ Kx> Kn < AB 20 1.301030 
 
 So is AB.AC ] .^ ,„^„ 
 
 \ AC 16.25 1.210853 
 
 To AP. AG - - - - 1.895802 
 
 AP = 8.50 0.929419 
 
 AG = 9.255 0.966383 
 
 PROBLEM 8. 
 Example 2. (PI. 2, fig. 11.) 
 
 Here, As BAC 100 ch. Ar. Co. 8.000000 
 
 Is to BDG 45 1.653213 
 
 So is BA^ S ^A 25 1.397940 
 
 (BA 1.397940 
 
 To BD^ - 2)2.449093 
 
 BD = 16.77 1.224546 
 
 PROBLEM 9. 
 
 Example 2. (PI. 2, fig. 12.) 
 
 Here the angles are, A= 71° 45', B = 49° 15', and C = 59°. Hence, 
 
 sin. A 71° 45' Ar. Co. 0.022414 
 
 As sin. A. sin. B , . t» ^r^o-.^, ^-io^m^ 
 
 \ sin. B 49° 15' " " 0.120580 
 
 T , J • r^ < radius 10.000000 
 
 Istorad.sm. C j ^.^^ ^ ^^^ _ _ _ _ ^^3^^^^ 
 
 So is 2 ABC 80 ch. 1.903090 
 
 To AB^ . . ^ 2)1.979150 
 
 AB = 9.763 ......... .989575 
 
52 SURVEYING. [Chap. IV, 
 
 PROBLEM 10. 
 
 Example 2. (PI. 2, fig. 13.) 
 
 Here the angles A == 99° 30', B = 122°, and P = 41° 30'. 
 
 , , , . . . T. ^ sin. A 99° 30' Ar. Co. 0.005997 
 
 And, As sm. A. sin. B | ^j^. ^ 122° - " " 0.071580 
 
 . ( radius 10.000000 
 
 Is to rad. . sm. P j ^^^^ p ^^o gQ, ... 9.821265 
 
 So is 2 ABCD 50 ch. - - 1.698970 
 
 To fourth term 39.61 - 1.597812 
 
 AB^ - - 36 "="""^ 
 
 CD^ - V 75.61 = 8.695. 
 
 Also, As sin. P 41° 30' Ar. Co. 0.178735 
 
 Is to sin. B 122° -...-.... 9.928420 
 
 So is DC— AB 2.695 0.430559 
 
 To AD 3.449 0.537714 
 
 Example 3. (PI. 3, fig. 1.) 
 
 Here the angles are, A = 90°, B = 73° 30', and P = 16° 30'. 
 
 . ( sin. A 90° - Ar. Co. 0.000000 
 
 Also, As sm. A.sm. B j ^j^^ g ^30 gQ, . » ,, 0.018263 
 
 ( radius 10.000000 
 
 Is to rad..sm. P j gj^ p jgo g^. . . . 9.453842 
 
 So is 2 ABCD 160 ch. .---.. . .'2.204120 
 
 To fourth term 47.39 1.675725 
 
 AB^ - 182.25 "^"""^ 
 
 CD = V 134.86 = 11.61. 
 
 And, As sin. P 16° 30' Ar. Co. 0.546658 
 
 Is to sin. B 73° 30' 9.981737 
 
 So is AB— CD 1.89 0.276462 
 
 To AD 6.38 0.804857 
 
Prob. 14.] LAYING OUT AND DIVIDING LAND. 
 
 PROBLEM 12. 
 
 Example 2. (PI 3, fig. 6.) 
 
 Here, As 2 : 1 : : 60^(100) : EF^ = 50, 
 
 EF = V50 = 7.07. 
 
 And, As BC (10) : EF (7.07) : : AB (15) ; AF = 10.605. 
 
 PROBLEM 13. 
 
 Example 2. (PI. 3, fig. 7.) 
 
 Here the angles are, A = 36^ 30', B = 100° 30', C = 43°, E 
 74° 30', and F = 69°. 
 
 As sin. E.sin. F 
 Is to sin. C . sin. B 
 So is BC^ 
 
 sin. E 74° 30 Ar. Co. 0.016089 
 
 sin. F 69 - - " " 0.029848 
 
 sin. C 43° 9.833783 
 
 sin. B 100° 30' - - - 9.992666 
 
 BC 18.66 1.270912 
 
 BC 1.270912 
 
 To fourth term 259.54 - - 2.414210 
 
 4 
 
 9)1038.16 
 
 EF = v/ 11 5.35 = 10.74. 
 
 As sin. A 36° 30' Ar. Co. 0.225612 
 
 Is to sin. E 74° 30' , 9.983911 
 
 So is EF 10.74- - - 1.031004 
 
 To AF 17.40 1.240527 
 
 PROBLEM 14. 
 Example 2. (PI. 3, fig. 8.) 
 
 YABM-JCD^^ 
 And, DC— AB (29.4): FE— AB (16.13) : : AD (30) : AF = 16. 16 
 
 Here, EF = \/[ " 2 "^ ' "^ \/ 5796. 18 = 76.13. 
 
54 
 
 SURVEYING. 
 
 [Chap. IV. 
 
 PROBLEM 16. 
 
 Example 1. (PL 3, fig. 12.) 
 
 The area of this tract may be found to be 858.552 square chains. 
 (The latitudes and departures are mostly given in the subsequent oper- 
 ations.) 
 
 To find area 
 
 ABODE, and the latitude and departure 
 
 of EA. 
 
 
 N. 
 
 s. 
 
 E. 
 
 w. 
 
 D. M. D. 
 
 N. Areas. 
 
 S. Areas. 
 
 AB 
 
 
 9.15 
 
 
 6.46 
 
 40.86 
 
 
 373.8690 
 
 BC 
 
 17.21 
 
 
 
 17.20 
 
 17.20 
 
 296.0120 
 
 
 CD 
 
 10.41 
 
 
 2.89 
 
 
 2.89 
 
 30.0849 
 
 
 DE 
 
 
 3.61 
 
 16.60 
 
 
 21.38 
 
 
 77.1818 
 
 EA 
 
 
 (14.86) 
 
 (6.17) 
 
 
 42.15 
 
 
 626.3490 
 
 
 27.62 
 
 27.62 
 
 23.66 
 
 23.66 
 
 
 326.0969 
 2ABCDEI 
 AEI 
 
 1077.3998 
 326.0969 
 
 
 751.3029 
 
 858.552 
 
 
 2)107.2491 
 
 
 63.6245 
 
 AE 
 
 14.J 
 
 w. 
 
 5.17 
 
 D. M. D. 
 
 5.17 
 
 N. Areas. 
 
 76.8262 
 
 S. Areas. 
 
 EF 
 
 FA 
 
 .93 
 
 21.49 
 
 21.49 
 
 19.9857 
 
 (15.79) 
 
 (16.32) 26.66 
 
 420.9614 
 
 96.8119 
 
 AEF 
 
 96.8119 
 
 2)324.1495 
 
 162.07475 
 
 As AEF 162.07475 Ar. Oo. 7.790285 
 
 Is to AEI 53.6245 1.729363 
 
 So is lat. EF .93 —1.968483 
 
 To lat. EI .31 —1.488131 
 
Prob. 16.] LAYING OUT AND DIVIDING LAND. 
 
 As AEF - • - Ar. Co. 7.790285 
 
 Is to AEI 1.729363 
 
 So is depart. EF 21.49 1.332236 
 
 To depart. EI 7.11 - - 0.851884 
 
 55 
 
 
 N. 
 
 s. 
 
 E. 
 
 w. 
 
 AE 
 
 14.86 
 
 
 
 5.17 
 
 EI 
 
 .31 
 
 
 7.11 
 
 
 lA 
 
 
 (15.17) 
 
 
 (1.94) 
 
 As diff. lat. 15.17 Ar. Co. 8.819014 
 
 Is to depart. 1.94 0.287802 
 
 So is radius - - 10.000000 
 
 To tang, bearing AI 7° 17' 9.106816 
 
 As COS. bearing .------ Ar. Co. 0.003518 
 
 Is to radius 10.000000 
 
 So is diff. lat. 15.17 1.180986 
 
 To dist. AI 15.29 1.184504 
 
56 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 - CHAPTER V. 
 MISCELLANEOUS QUESTIONS. 
 
 Question 1. 
 
 square yarc 
 
 / / 2420 \ 
 And radius = \/ (^rnTT^) = v/ 770.3081 = 27.75. 
 
 Here 4 Acre = 2420 square yards ; 
 2420 \ 
 
 Question 2. 
 
 Construction. 
 
 Make AB (PI. 4, fig. 1) = 40 = one of the given sides, and at A 
 
 320 
 draw AL perpendicular to AB and = -rjr = 8 ; through L draw GH 
 
 parallel to AB, and with the centre A and distance = 20 = the other 
 given side, describe an arc, cutting GH in D and C ; join AC, BC, 
 AD, and BD : then will ABC and ABD answer the conditions of the 
 pestion. 
 
 Calculation. 
 
 AE = AF = VAC^ — CE^ = ^400 — 64 = 18.3303 ; and BE = 
 AB— AE = 21.6697 ; therefore, BC= v/ BE" + EC^=: V 533.57589809 
 = 23.099. Also, BF = AB + AF= 58.3303, and BD = n/BF+FP 
 
 = V 3466.42389809 = 58.876. 
 
 Another Solution. 
 
 Find AE = AF as before. Then, from Geometry, BC' = AB' + 
 AC^ - 2 AB.AE = 2000 - 1466.424 = 533.576, and BC = 23.099. 
 Also, BD^ = AB^ + AD^ + 2 AB.AF - 2000 + 1466.424 = 3466.424, 
 and BF ^ 58.876. 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 57 
 
 Question 3. 
 
 Here it is evident the number of acres will be inversely as the 
 number of square yards in a Perch : 
 
 Therefore, 6^ : 5.5^ :: 110 A. : 92 A. 1 R. 28^ P. Cheshire. 
 And T : 5,5^ :: 110 A. : 67 A. 3 R. 25 if P. Irish. 
 
 Question 4. 
 
 28 7 
 Here-r7r = — = twice the thickness of the wall, also 840 links = 
 
 554.4 feet=the longer diameter within the walls ; 612 links=:403.92 
 
 , , 7 1670.2 , ,. ., , 
 
 feet — the shorter; 554.4 + -= — - — =: longer diameter outside, and 
 
 7 1218 7fi 
 
 403.92 +^= ^ = shorter. By Frob. 10, Chap. III. the area 
 
 o o 
 
 within the walls = 554.4 x 403.92 x .7854 = 223933.248 X .7854 = 
 
 175877.1729792 ft. = 4 A. R. 6 P. The area to the outside = 
 
 1670.2 1218.76 2035572.952 1598738.9965008 
 -^— X X .7854 = X .7854= ^ 
 
 = 177637.6662778 feet. Therefore 177637.666 — 175877.173== 
 1760.493 = area the wall stands upon. 
 
 Question 5. 
 
 Here the area of an ellipse whose diameters are 3 and 2 is 4.7124. 
 Then, since similar figures are as the squares of their like dimensions, 
 we have. As 4.7124 : 160 :: 9 : 305.5768 = square of the longer dia- 
 meter; consequently V 305.5768 = 17.481 = longer diameter; and 
 3:2:: 17.481 : 11.654 = shorter diameter. 
 
 Question 6. 
 
 Find the area of the triangle whose sides are 9, 8, and 6 ; thus, 
 
 ' ^ ' -= 11.5, and 711.5x2.5x3.5x5.5= 7553.4375 =23.525 
 
 square perches. Also, 6 A. 1 R. 12 P. = 1012 P., and 23.525 : 1012 
 :: 82 : 2753.1562 = square of the second side; therefore 72753.1562 
 = 52.47. Also, 
 
 8:9:: 52.47 : 59.029 = longest side. 
 
 8:6:: 52.47 : 39.353 = shortest side. 
 
58 
 
 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 Question 7. (PI. 4, fig. 2.) 
 To find ABC : 
 
 AB 27.35 
 
 EC . 31.15 
 
 CA 38.00 
 
 2)96.50 
 
 Half sum 48.25 ' 1.683497 
 
 r 20.90 1.320146 
 
 Remainders j 17.10 1.232996 
 
 ( 10.25 -.-.... 1.010724 
 
 2)5.247363 
 
 ABC 420.417 2.623681 
 
 To find ACE: 
 
 AC 38. 
 
 CE 40.10 
 
 EA 22.20 
 
 2)100.30 
 
 Half sum 50.15 - - 1.700271 
 
 ( 12.15 -...-.. 1.084576 
 
 Remainders } io.05 1.002166 
 
 (27.95 1.446382 
 
 2)5.233395 
 
 ACE 413.71 ..-....-.. 2.616697 
 
 To find CED: 
 
 CE 40.10 
 
 CD 23.70 
 
 DE 29.25 
 
 2)93.05 
 
 Half sum 46.525 1.667686 
 
 C 6.425 0.807873 
 
 Remainders j 22.825 -.-.--. 1.358410 
 
 (17.275 1.237408 
 
 2)5.071377 
 
 CED 343.311 2.535688 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 59 
 
 Hence the whole area = 420.418 + 413.71 + 343.308 = 1177.436 
 Ch. = 117 A. 2R. 38.976 P. 
 
 Question 8. 
 Construction. 
 Make AB (PI. 4, fig. 3.) equal to half the given perimeter = 52, 
 and bisect it in D ; make DC perpendicular to AB and equal to the 
 square root of the given area ; with the centre C and radius equal 
 to AD, describe an arc cutting AB in E, complete the rectangle 
 AEFG and it will be the one required. The demonstration is evi- 
 dent from Geometry. 
 
 Calculation. 
 
 DE = v/CE^— CD^ = ^/ 676— 480 = n/196= 14. 
 
 AE = AD + DE = 26+14 = 40, and EF = EB = 26—14 = 12. 
 
 Question 9. 
 Construction. 
 Draw any line AC, (PI. 4, fig. 4.) and in it take AE = 20 = given 
 difference ; make EF perpendicular to AC = 20 ; join AF and pro- 
 duce it to B, making FB = 20 ; then will AB be a side of the square. 
 
 Demonstration. 
 Since EA =:EF,the angles FAE and AFE are each equal to half 
 a right angle, and AC must be the diagonal of the square. Again 
 the triangles CEF and CBF are equal, since they are right angled 
 at E and B, and have the hypothenuse and one leg in each equal : 
 we have therefore CE = CB = CA— 20. 
 
 Calculation. 
 
 AF =VAE'' + EF2= n/800 = 28.284, and AB = AF + FB = 
 48.284 ; hence the area = AB^ = 2331.344656 sq. per. = 14 A. 2 R. 
 11.34 P. 
 
 Question 10. 
 Construction. 
 Let ABCD (PI. 4, fig. 5.) be the given rectangle. In BA and BA 
 produced take BH = BC, and AR = I AD. On BR describe the 
 semicircle BPR, meeting DA produced in P ; bisect AH in O, and 
 with the centre O and radius OP, describe the semicircle EPQ, 
 make AG = AQ, complete the rectangle AF, and the thing is done. 
 
60 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 Demonstration. 
 
 AF = AEx AG = AEx AQ = AP^ = ABx AR = f ABx AD = 
 t AC. Also, BE = BH— HE = BC— AQ = AD— AG = GD. 
 
 Calculation. 
 AO = 1 AH = 10 : AP^ = ABx f AD = 6000 ; therefore, OP = 
 
 v/AP^ + OA^ = V6100 = 78.1025; BE = BO— OE = 90—78.1025 
 = 11.8975. 
 
 Question 11. 
 
 Construction. 
 Let ABD (PI. 4, fig. 6.) be the given circle. Draw the diameter 
 AB and radius CD perpendicular to it ; take CF = | AC ; upon BF 
 describe a semicircle cutting CD in E : with C as a centre and 
 radius CE, describe the circle EGH, and the thing is done. 
 
 Demonstration. 
 
 CE is a mean proportional between CF and CB ; hence CF : CB : : 
 CE^ : CB^ : : 4 : 5 ; and since circles are as the squares of their radii, 
 we have GEH = j ABD. 
 
 Calculation. 
 
 V5 : v^4 : : AC (75) : EC = Z^^ 
 
 >/5 
 
 ^150vA5 ^ 3Q^g ^ 67.082, and DE = DC— EC = 7.918. 
 5 
 
 Question 12. 
 
 Construction. 
 
 With the given distances form the triangle ABC, (PI. 4, fig. 7.) 
 
 Upon AB describe the equilateral triangle ABD ; join CD and on it 
 
 describe the equilateral triangle CDE, which will be the one required. 
 
 Demonstration. 
 
 Since BD and BC are by construction two of the given distances ; 
 it is only necessary to prove that BE = AC, which is evident from 
 the equaUty of the triangles DAC and DBE. 
 
Chap.V.] miscellaneous questions. 61 
 
 Calculation. 
 In the triangle ABC, find the angle BAG, thus, 
 BC 10 
 
 AC 7.5 Ar. Co. 9.124939 
 
 AB 12.5 ....... Ar. Co. 8.903090 
 
 2)30. 
 
 15 1.176091 
 
 5 0.698970 
 
 2)19.903090 
 
 Cos. i BAC 26° 34' - . - - . 9.951545 
 
 BAC 53° 8' 
 
 Then in the triangle DAC we have DA and AC, and the angle 
 DAC = 113° 8' to find DC, thus. 
 
 As DA + AC 20 - Ar. Co. 8.698970 
 
 Is to DA— AC 5 0.698970 
 
 So is tang. — A±^5^ . 33° 26' .... 9.819684 
 
 ^ 2 
 
 To tang. DCA— ADC _ go 33- .... 9.217624 
 
 ACD - 42° 48' 
 And, 
 
 As sin. ACD 42° 48' Ar. Co. 0.167848 
 
 Is to sin. DAC 113° 8' - 9.963596 
 
 So is AD 12.5 - - 1.096910 
 
 To DC 16.92 1.228354 
 
 Then in CDE, we have the sides and angles to find the area 
 
 thus. 
 
 As radius Ar. Co. 0.000000 
 
 Is to sin CDE 60° 9.937531 
 
 o • nr^v^rkT? (CD 1.228354 
 
 SoisCDxDE jj^^ ^228354 
 
 To 2 CDE 247.88 2.394239 
 
 123.94 Ch. = 12 A. 1 R. 23.04 P. 
 
62 
 
 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 Question 13. 
 
 Construction. 
 With the given bearings and distances protract the figure ABCDfg 
 PL 4, fig. 8. Join Ag, and with the centres g and A, and distances 
 equal to the 4th and 7th sides, describe arcs cutting in G. Join AG 
 and gG, and draw DE, EF, and FG respectively parallel and equal 
 to gG, Df, and fg. Then will ABCDEFG be the required map. 
 
 Calculation. 
 To find the bearing and distance of gA. 
 
 
 Bearing. 
 
 Dist. 
 
 N. 
 
 S. 
 
 E. 
 
 W. 
 
 AB 
 
 S. 72W. 
 
 24.00 
 
 
 7.42 
 
 
 22.83 
 
 EC 
 
 North. 
 
 38.00 
 
 38.00 
 
 
 
 
 CD 
 
 N. 824 E. 
 
 41.00 
 
 5.35 
 
 
 40.65 
 
 
 Df 
 
 S. 80 E. 
 
 11.50 
 
 
 2.00 
 
 11.32 
 
 
 % 
 
 S. 26 W. 
 
 22.00 
 
 
 19.77 
 
 
 9.64 
 
 gA 
 
 
 
 
 (14.16) 
 
 
 (19.50) 
 
 
 
 
 43.35 
 
 43.35 
 
 51.97 
 
 51.97 
 
 As diff. lat. 14.16 Ar. Co. 8.848937 
 
 Is to departure 19.50 - - - - ■- 1.290035 
 
 So is radius 10.000000 
 
 To tang, bearing gA S. 54° V W. 
 
 10.138972 
 
 As COS. bearing 54° 1' --.--. Ar. Co. 0.230955 
 
 Is to radius - - - - 10.000000 
 
 So is diff. lat. 14.16 1.151063 
 
 To distance gA 24.10 
 
 1.382018 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 63 
 
 In the triangle AGg we have the sides to find the angles AgG and 
 
 GAg; 
 
 Thus, 
 
 AG 37 
 
 gG 20 Ar. Co. 8.698970 
 
 Ag 24.1 Ar. Co. 8.617982 
 
 2)81.1 
 
 Half sum 40.55 1.607991 
 
 » 
 
 Remainder 3.55 0.550228 
 
 2)19.475171 
 Cos. i AgG 56° 52i' - - - - 9.737585 
 AgG 113° 45' 
 
 And, 
 
 As AG 37 -..----. - Ar. Co. 8.431798 
 
 Is to gG 20 1.301030 
 
 vSo is sin. AgG 113° 45'.- - - .... 9.961569 
 
 To sin. gAG 29° 39' - - 9.694397 
 
 Applying now the bearing of gA to these angles we will have the 
 bearing of gG or DE = S. 59° 44' E, and of GA = S. 83° 40' W 
 The area will then be calculated as in the following table, viz. 
 
64 
 
 
 MISCELLANEOUS QUESTIONS. 
 
 [Chap. V. 
 
 
 
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 Tt< 
 
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Chap, v.] MISCELLANEOUS QUESTIONS. 65 
 
 Question 14. 
 Construction. 
 Make AB, (PI. 4, fig. 9.) = the given side, and divide it in D, so 
 that AD naay be to DB in the ratio of 3 to 2 ; in AB produced, take 
 DO a fourth proportional to AD — DB, DB, and AD, and with the 
 eentre O and radius OD, describe the semicircle DCE ; make AG 
 perpendicular to AB, and equal to tw^ice the area divided by AB = 
 6 ; through G draw GF parallel to AB, cutting the circle in C and 
 F ; join AC BC, AF and BF ; then will ABC and ABF answer the 
 conditions of the question. 
 
 Demonsti^alion. 
 Since AD— DB : DB : : AD : DO, we have AD : DB : : AO : DO 
 or AO : AD : : DO : DB, therefore, AO : DO : : DO : OB, con- 
 sequently (Euclid, F. 6.) AC : BC : : AD : DB : : 3 : 2; and AF : BF 
 : AD : DB : : 3 : 2. 
 
 Calculation. 
 
 As 3 + 2 : 15 : : 3 : AD = 9, and DB = 6; also, 9—6 : 6 : : 9 : DO 
 
 = 18, and AO = 9 + 18 = 27, join OC, and OF, and let fall the per- 
 pendiculars CL and FP ; then OL = /OC"— CL^ = V324— 36 = 
 v288 = 16.9706, and AL = AO— OL = 10.0294; hence AC = 
 /AL^ + LC"= n/ 136.58886436 = 11.6871; and as 3 : 2 : : 11.6871: 
 BC = 7.7914. Again AP = AO + OP = 43.9706, and AF = 
 
 ^/'AP + PF = y 1969.41366436 = 44.3781 ; and as 3 : 2 : : 44.3781 : 
 BF = 29.5854 
 
 Question 15. 
 Construction. 
 Make AB, (PI. 4, fig. 10.) = the given side, and BL = the sum 
 of the other sides ; Bisect AB in D, and take DH a third propor- 
 tional to 2 AB and'BL ; Draw HE perpendicular to BH and equal 
 
 to — — - = 32. Through E draw EF parallel and equal to BL ; 
 
 join EA and produce it to G, making FG = AB ; draw AC parallel 
 to FG, and join BC ; then ABC is the triangle required. 
 
 Demonstration. 
 By Construction BL" = 2 AB x DH ; also, in the similar triangles 
 EGF and EAC, we have GF (AB) : AC : : EF (BL) : EC (HP). 
 
 5 
 
66 MISCELLANEOUSQUESTIONS. [Chap. V. 
 
 Hence BL x AC = GF x HP, or 2 BL x AC = 2 GF x HP. Subtracting 
 these equals from the precedmg, we have BL^ — 2BLxAC = 
 2ABxDH — 2GFxHP = 2ABxDP = (BP + AP)x(BP — AP) = 
 BP2_AP2 = BC^— AC^. Hence BL/— 2 BL x AC + AC^ = BC^ and 
 BL — AC = BC, or BL ^ BC + AC. 
 
 Cdlculation. 
 
 As 2 AB (100) : BL (85) : : BL (85) : DH = 72.25, and AH = 
 DH — AD = 47.25. Now m the right angled triangle AHE, we 
 have the sides AH and HE, to find HAE and AE ; thus, 
 
 As AH 47.25 ------- Ar. Co. 8.325598 
 
 Is to HE 32 -------- - - 1.505150 
 
 So is radius 10.000000 
 
 To tang. HAE 34° 6 i' - -. - - - - - 9.830748 
 
 And, 
 
 As COS. HAE 34° 6^ - - - - - Ar. Co. 0.081981 
 
 Is to radius 10.000000 
 
 So is AH - - 1.674402 
 
 To AE 57.07 - - - ^ 1.756383 
 
 Now in the triangle GEF we ha\'c FE, FG, and the angle FEG 
 = HAE, to find FGE ; thus, 
 
 As FG 50 - Ar. Co. 8.301030 
 
 Is to FE 85 1.929419 
 
 So is sin. GEF 34° 6 ^ 9.748776 
 
 To sin. FGE 72° 25' ------- - 9.979225 
 
 Finally, in ACE we have AE and the angles to find AC ; thus, 
 
 As sin. ACE 73° 284' - - - - Ar. Co. 0.018319 
 
 Is to sin. AEC 34° 6i' - - ' - - - - - - 9.748776 
 
 So is AE - - - - 1.756383 
 
 To AC 33.3793 1.523478 
 
 And BC = 85 — 33.3793 = 51.6207. 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 67 
 
 Question 16. 
 Construction. 
 Make AC (PI. 4, fig. 11.) = 50 = the given diagonal, and on itdescriV-e 
 
 a semicircle ABC ; make AE perpendicular to AC and = ■- = 
 
 24 ; draw EB parallel to AC, cutting the semicircle in B ; join AB, BC, 
 and draw CD and DA parallel to them; then will ABCD be the 
 rectangle required. 
 
 Demonstration. 
 
 Since ABC is an angle in a semicircie, it is right, and ABCD is <. 
 rectangle. Also its area = AC x BF = 1200 perches = 7i acres. 
 
 Calculation. 
 
 FG = v/BG^— BF^' r= V49 = 7 ; AF = AG— GF = 18, and AB 
 = VAF' + FB^ = V900 = 30, BC = x/AC^— AB^ = V1600 = 40. 
 
 QuESTIOIV 17. 
 
 Construction. 
 Make AB (PI. 4, fig. 12.) = the square root of the given aira, and 
 draw CE perpendicular to it : draw BC, making ABC = 30°, make 
 AE = AC ; bisect AC in D, and draw EF perpendicular to CE and 
 = ED. Complete the parallelogram CEFG, which will be the one 
 required. 
 
 Demonstration. 
 Since the angle B = 30°, and A = 90°, BC = 2 AC = Cl!i = 4 CD, 
 and EF = ED = 3 CD ; therefore FC = VEF^ + EC^ =- 5 CD. Also 
 AB^ = BC^ — AC^ = IBC^ = lEC^ = EC x ED ^ EC> EF = 
 CEFG. . 
 
 " Calculation. 
 
 Since AB^ = | CE' CE^ = f AB^ = f the given area = 784, and 
 CE = 28 ; hence EF ■- | EC = 21. 
 
 Question 18. 
 
 Construction. 
 With the given bearings and distances protract the figure ABCD, 
 (PI. 4, fig. 13.) and from B draw BP according to the given bearing 
 
68 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 and distance of the spring. Produce DA and CB to meet in F, and 
 through P draw EH parallel to AD. Bisect AF in G, join EG, and 
 draw BM parallel to it, and MN parallel to FE. Make MT perpen- 
 dicular to MN, and equal to the square root of the given area. Take 
 MU a third proportional to MN and MT ; draw UH parallel to MN, 
 cutting AF in I; draw IK perpendicular to AF and equal to EP, and 
 with the centre K and distance PH describe an arc cutting AD in 
 Q ; draw QPR, and the thing is done. 
 
 Denumstratiov . 
 
 In the similar triangles FGE and FMB, we have FB : FM 
 : : FE : FG ; therefore, 15.6, the triangle EFM = BFG ; but EFM= 
 1 FMNE, and BFG = h BFA ; hence FMNE = BFA. Again, be- 
 cause the triangles EPR, IQS, and PHS are similar, and the homo- 
 logous sides EP (IK), IQ, and PH (KQ) form a right angled trian- 
 gle, we have from Geometry EPR + IQS=SPH. Add FISPE to 
 each, and we have FQR = EFIH. But FBA = EFMN, hence 
 BAQR = MNIH = MN.MU =MT2 = the given area. 
 
 Calculation. 
 
 From the bearings of the lines the angles may be found as follow. 
 AFB = BEP = 23°, ABF = 84° 30', BAF = 72° 30', EBP - 145° 
 30', and EPB = 11° 30'. Then, in the triangle EBP we have all the 
 angles and side BP, to find EP and EB : 
 
 Thus, 
 
 As sin. BEP 23° - - Ar. Co.. 0.408122 
 
 Is to sin. EBP 145° 30' - - 9.753128 
 
 So is BP 7.90 0.897627 
 
 To EP 11.452 1.058877 
 
 And, 
 
 As sin. BEP Ar. Co. 0.408122 
 
 Is to sin. BPE 11° 30' 9.299655 
 
 So is BP 0.897627 
 
 To BE 4.031 0.605404 
 
Chap.V]. miscellaneous questions. 69 
 
 Also, in the triangle ABF, all the angles and side AB are given, 
 tofindBFand AF-; 
 
 Thus, 
 
 As sin. AFB 23° Ar. Co. 0.408122 
 
 Is to sin. BAF 72° 30' 9.979420 
 
 So is AB 15.20 1.181844 
 
 To BF 37.101 - - 1.569386 
 
 And, 
 
 As sin. AFB Ar. Co. 0.408122 
 
 Is to sin. ABF 84° 30' 9.997996 
 
 So is AB - . - - 1.181844 
 
 To AF 38.722 1.587962 
 
 And FE = FB— BE = 33.07, and FG = i AF = 19.361 ; 
 Also, 
 
 As EF 33.07 .--..--- Ar. Co. 8.480566 
 
 Is to GF 19.361 1.286927 
 
 So is BF 37.101 "1.569386 
 
 To FM 21.721 1.336879 
 
 Now, in the parallelogram MIHN, we have MN = FE = 33.07, 
 and IMN = F = 23°, and the area = 100 square chains, to find MI; 
 Thus, 
 
 MN 33.07 - Ar. Co. 8.480566 
 
 As MN, sin. IMN ^ ^.^^ j^^ ^3, ^^ ^^^ ^^^^^^2 
 
 Is to radius 10.000000 
 
 So is MIHN 2.000000 
 
 To MI 7.739 - - '- 0. 
 
 Therefore PH = EH — EP = FM+MI — EP = 18.008. Now, 
 in the right angled triangle IKQ, we have IK = EP = 11.452, and 
 KQ = PH = 18.008, to find IQ; thus, 
 
 KQ+KI = 29.46 log. 1.469233 
 KQ— KI = 6.556 0.816639 
 
 2)2.285872 
 
 IQ= 13,898 1.142936 
 
 Hence AQ = FQ — FA = FM+MI + IQ — FA = 4.636. 
 
70 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 Question 19. 
 
 Construction. 
 
 With the given bearings and distances, protract the figure ABCD, 
 (PL 4, fig. 14;) then, by Prob. 15, Chap. IV. divide ABCD into two 
 equal parts by the Hne EF, parallel to CD ; also, by the same pro- 
 blem, divide ABCD, and EBAF, each into two equal parts by the 
 lines OM and PN, parallel to AD ; join MN, produce it to I, and 
 draw OH parallel to IM; join IH, then will EF and IH be the 
 division hnes required. 
 
 Demonstration. 
 
 Because PN is parallel to OM, we have IN : NM : : IP : PO 
 : : IG : GH, because NG is parallel to HM ; therefore, PG is parallel 
 to OH, and consequently to IM. Now since OH is parallel to IM, 
 we have IHM = lOM, to each add AIMD, and AIHD = AOMD = 
 \ ABCD. In the same manner it may be shown that AIGF = 
 i ABEF = i ABCD. 
 
 Calculation. 
 
 Draw EK and IL, each parallel to AD, and MU parallel to AB. 
 From the given bearings find the angle A = 78° 30', B = 139° 45', 
 C = 78° 45', and D = 63°. By Prob. 15, Chap. IV., find EF and 
 AF, thus, 
 
 . • r^ ■ T^ \ sin- C 78° 45' Ar. Co. 0.008426 
 
 Is to sin. A. sin. B j .-"• A 78° 30' - - - .- 9.991193 
 i sin. B 139° 45' ... 9.810316 
 
 SoisAB^ j^B 23 ^-3^1^28 
 
 ^ AB 1.361728 
 
 To fourth term 383.274 2.583510 
 
 CD" 2161.3201 . 
 
 2)2544.5941 
 EF = 1/1272.2970 = 35.67 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 
 
 And in the triangle ECK, 
 
 As sin. E 38° 15' Ar, Co. 0.208243 
 
 Is to sin. C 78° 45' 9.991574 
 
 So is CD— EF 10.82 - 1.034227 
 
 To FD - 17.14 - - -1.234044 
 
 AD 49.64 
 
 AF 32.50 
 
 71 
 
 Then in the triangle ECK, we have the angles and side EK = 
 FD, to find EC, thus, 
 
 As sin. C 78° 45' - - • - - - - Ar. Co. 0.008426 
 
 Is to sin. K 63° - - ' 9.949881 
 
 So is EK 17.14 1.234044 
 
 To CE 15.57 1.192351 
 
 Consequently BE = BC — CE = 14.93. Now by the same pro- 
 l>lem find OM, AO, FN and AP, thus, 
 
 . A • -n i ^^"- A - - Ar. Co. 0.008807 
 
 As sin. A. sin. D ^ ^.^^ ^ _ _ ^^_ ^^_ 0.050119 
 
 . „ . ^ ( sin. B 9.810316 
 
 Istosm.B.sm.C j ^.^^ ^ ...... 9.991574 
 
 ( BC 30.50 ... - 1.484300 
 
 ^^ *^ -^^^ i BC 1.484300 
 
 To fourth term 675.18 2.829416 
 
 AD^ 2464.1296 == 
 
 2)3139.3096 
 OM = V 1569.6548 = 39.62 
 
 And, 
 
 As sin. EMD 38° 30' Ar. Co. 0.205850 
 
 Is to sin. D 9.949881 
 
 So is AD— OM 10.02 1.000868 
 
 To AO 14.34 1.156599 
 
72 MISCELLANEOUS QUESTIONS. [Chap. V. 
 
 And, 
 
 l Sin. F - - - Ar. Co. 0.050119 
 
 . „ . „ ( sin. B 9.810B16 
 
 Is to sm. B.sin. ii J • r, nnn-i^^nA 
 
 ( sin. E 9.991574 
 
 . (BE 14.93 .... 1.174060 
 So is BE j gj, ^^^^^g(^ 
 
 To Fourth term 161.78 2.208936 
 
 AF^ .... 1056.25 
 
 2)1218.03 
 PN ... ^/609.01= 24.68. 
 
 And, 
 
 As sin. HMD 38° 30' . . - - Ar. Oo. 0.205850 
 
 Is to sin. F 9.949881 
 
 So is AF — PN 7.82 ........ 0.893207 
 
 To AP 11.19 1.048938 
 
 Hence OP = AO — AP = 3.15 ; wherefore we have 
 
 OM— PN (14.94) : PN (24.68) : : OP (3.15) : IP = 5.20 ; and AI = 
 AP— IP = 5.99. 
 
 In the triangle MUL we have the angle U = A, L = D, and side 
 MU = 10 = IP + PO == 8.35, to find ML and UL; thus, 
 
 As sin. ULM 63° Ar. Co. 0.050119 
 
 Is to sin. MUL 78° 30' .... . ■ . 9.991198 
 So is MU 8.35 ..." 0.921686 
 
 To ML 9.18 0.962998 
 
 And, 
 
 As sin. MLU 63° Ar. Co. 0.050119 
 
 Is to sin. UML 38° 30' 9.794150 
 
 So is MU 8.35 0.921686 
 
 To UL 5.83 0.765955 
 
Chap, v.] MISCELLANEOUS QUESTIONS. 73 
 
 Therefore IL = lU+UL = OM+UL = 45.45, and from the simi- 
 lar triangles ILM and OMH, we have 
 
 As IL (45.45) : LM (9.18) :: OM (39-62) : MH = 8. 
 
 Now, in the triangle ILH, we have the angle L = D, and sides 
 [L and LH = LM4-MH = 17.18, to find the angle LIH; thus, 
 
 As LI + LH G2.63 Ar. Co. 8.203218 
 
 Is to LI — LH 28.27 - - 1.451326 
 
 So is tang. LHI + LjH _ ^go 30' . . . . io.212681 
 
 LIH - ... 22 7 
 Bearing IL - - 66 15 
 
 IK - S. 88 22 E. 
 
 To tang. i^5? yi? . 36 23 - - - - 9.867225 
 
 * 2 =^=^= 
 
I 'I.. in: I 
 
I'lJIT. 1 
 
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