(4.) Sin. (60°+A)- sin. (60°-A)=2 cos. 60° sin. A (60)= sin. A. For by (Art. 84) cos. 60°, hence 2 cos. 60°=1. Again, cos. (30°-A)-cos. (30+ A)=2sin. 30° sin. A. (62)= sin. A. For by (Art. 84) sin. 30°, hence 2 sin. 30°=1. (6.) Article 36 gives == hence (Alg. 108) and b sin. A hence sin. A+ sin. B = a+b sin. Asin. B = a-b sin. A--sin. B' tan. (AB) sin. Asin. B tan. (A-B) -b tan. (A-B)' (59.) and since a and b are the sides opposite to the angles A and B, the expression proves, that the sum of the sides is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference, which is the rule. (7.) Let (AD-DC)=d then, c―a: d=p cosec. A-p cosec. C: p cot. A-p cot. C (÷p). =cosec. A- -cosec. C: cot. A- -cot. C(× sin.A sin. C). sin. C-sin. A: sin. C cos. A-cos. C sin. A. sin. C-sin. A: sin. (C-A) (53. =2sin.(C-A)cos. (C+A):2sin. (C-A)cos.§(C−A) (60 and 69).) =cos.(C+A): cos.(C-A). (÷2 sin.C-A). Again, c+a: d=p cosec. A+p cosec. C:p cot. A-p cot. C. =cosec. A+cosec. C: cot. A-cot. C (÷p). sin. C+sin. A: sin. C cos. A-cos. C sin. A. 2 sin. (C+A) cos. § (C-A): 2 sin. § (C—A) cos. & (C—A). sin. (CA): sin. (C-A) (÷2 cos. C-A). Q. E. D. (2.) 206.148 Art. 87. 160: the height above the eye. = 10.000000 =10.099395 2.204120 2.303515 height of the eye. R: tan. 47° 30' 100: the height above the eye. (3.) Log. R Log. tan. 47° 30' Log. 100 10.000000 10'034948 tower. 2.000000 2.037948 R: tan. 50° 43′=130: height above the eye. (1.) Sin. 131° 26', or cos. 41° 26': sin. 38° 40′ 266 : the ne dist and Sin. 131° 26': sin. 92° 46′ 266: the other distance. Ar. co. Log. sin. 131° 26'-Log, cosec. 131°26′ =10.125097 Log. sin. 38-40 Log. 221-672 = 9.795733 -2-424882 = 2.345712 Ar. co. Log. sin. 131° 26'-Log. cosec. 131°26′ =10-125097 Log. 354-382 9.999493 2.424882 = 2.549472 (2.) NE. N. is 3 points to the east of the north, and NW. W. is 44 points to the west of the north, the sum of these, or 7 points, is the angle formed by the ship's course, and the bearing of the headland. Again, the course the ship had when at the second station was SE. E., which is just 4 points from the bearing of the headland: hence the angles at the extremity of the base were 73 points and 4 points, the sum of which is 11 points, and since the base is given 20 miles, we have C Sin. 44 pts.: sin. 4 Sin. 4 pts.: sin. 7 pts.=20: Dist. from first station. Ar. co. Log. sin. 44 pts. Log. cosec. 44 pts. Log. sin. 4 pts. Log. 20 Log. 19-0864, or 19.09 nearly 10.130210 = 9.849485 1.301030 ar. co. Log. sin. 44 pts.=Log. cosec. 44 pts. 10-130210 Log. sin. 7 pts. Log. 20 Log. 26.96 = 12-80725 = 9.999477 = 1.301030 = 1.430717 (3.) Sin. 60°: sin. 80°-260 : side op. 80°. obtain perp. breadth= 260xsin. 80° sin. 40° sin. 60° breadth. =260 × sin. 80° sin. 40% cosec. 60°; hence Log.cosec. 60° Log.' sin. 40° Log. sin. 80° Log. 260 Log. 190-046 =2.278860 Art. 89. 180°-174° 13′=5° 47′ the angle directly under the top of Arthur's Seat subtended by the distance of the stations, on the same horizontal plane as the base. 3.259 correction for curvature. 821.174 height above the medium level of the tide. 4.067220 2 8.134440 Constant 8.378641 =0.513081 Constant 8.378641 2.808 correction for curvature. 821-019 height above the medium level of the tide. = 448417 (1.) In ABCD there are given LBCD, 37°; LBDC= LADB+LADC; :. [BDC=96°; hence the third angle CBD 47°, and the side CD=300 yards, to find ČB and DB. = } By Art. 36. 10.135873 = 9-779463 Sin. CBD: sin. BDC-CD: CB = 2.477121 LACD LACB+LBCD; .. LACD=93°; hence the third angle CAD=46°, and the side CD=300 yards, to find AC and AD. Sin. CAD: sin. ACD=CD: AD Sin. CAD: sin. ADC=CD : AC ƒ By Art. 36. 10-143066 = 9.816943 Log.cosec. 46° 10-143066 Log.cosec. 46° Log. sin. 93° = 9.999404 Log. sin. 41° Log. CD 300 = 2.477121 Log. CD, 300, = 2.477121 Log. AD416-477= 2.619591 Log.AC, 273-609= 2·437130 Next, in the AACB there are given the side AC= 273 609, and CB=407·95, and the contained [ACB=56o, to find AB (180°—56°)—26°, which is therefore half the sum of the angles at A and B. Hence by Art. 40, CB=407.95 AC=273.609 Ar. co. Log. 681-559, sum of sides, 17-166496 Log. 134-341, dif. of sides, = 2.128208 Log. tan. 62° =10.274326 Log. tan. 20° 20' 24" = 9.569030 LCAB 82° 20′ 24′′ LCBA 41° 39' 36" Lastly, to find the side AB from ABC. Log. cosec. CAB=82° 20′ 24′′-10-003893 Log. sin. 56° Log. CB=407.95 Log. AB=341-252 =9-918574 = 2.610608 =2.533075 The same by the method of rectangular co-ordinates. |