of c which are obtained from the two values of A in the equality sin A = sin B (see last Art.). . And the angles ACB and A'CB correspond to the two values of C which would also be found. Cor. If a perpendicular CD be drawn from C upon AA, and if c' and c be the lengths of BA' and BA respectively, it may be easily shown that c' + c = 2 a cos B, and c' c = 2 bcos A. 4. a = Ex. VIII. Solve the following triangles, having given1. b = 12, c = 6, A = 60°. 2. a = 18, b = 18 J2, A = 30°. 3. a = 5 13,6 = 5 V2, c = }( 16 + /2). 12, B 60°, C = 15°. Log 121 = 2.0827854, L cot 20° 16' = 10.4326795, 2.7944880, L tan 27° 44' = 9.7207827, 237, 6 341, B 28° 24', find A. 2:5327544, L sin 19° 18' = 9.5191904, Log 2 = 3010300, L cot 13° 16' = 10.6275008, 16, c = 18, find A, B. :3010300, log 3 •4771213, 9.6506199, L tan 29° 12' = 9.7473194, L tan 29° 13' = 9.7476160. 8. a = Log 341 10. a = 14, 6 log 2 Log 3 11. a = 3, 6 2 A 60°, find B, C, and c. Log 2 = -3010300, log 3 = 4771213, Log 1.3797 = 1397847, log 1.3798 = .1398161. •4771213, L tan 22° 13' 9.6111196. 13. If c, c' be the two values of the third side in the ambiguous case when a, b, A are given, show that (c – c)2 + (c + c') tano A = 4.a”. 14. If a, b, A are given, show from the equation 62 + ca 2 bc cos A = aʼ; that if c and c' be the two values of the third side cc' 62 - a', and c + c 26 cos A. 15. Show also from the same equation that there is no ambiguous case when a = b sin A, and that c is impossible when a <b sin A. 16. Having a - b, A, B, solve the triangle. 17. Given the ratios of the sides, and the angle A, solve the triangle. A } – C? 2, 6 COS P = C. CHAPTER X. HEIGHTS AND DISTANCES. 43. We shall now show how the principles of the previous chapters are practically applied in determining heights and distances. We have not space here to describe the instruments by which angles are practically measured, but we shall assume that they can be measured to almost any degree of accuracy. 1A any 'C tan 0. a 200. 44. To find the height of an accessible object. Let AC be the object, and let distance BC from its foot be measured. At B let the angle of elevation ABC be observed. Suppose BC = a, 2 ABC = 0. Then, we have AC BC :: AC = a tan o, the height required. Ex. Let a = 200, and 0 = 30°. 1 200 13. Then AC 200 tan 30° V3 3 45. To find the height of an inaccessible object. At any point B in the horizontal plane of the base let the angle of elevation ABC be observed. Measure a convenient distance BD in the straight line CB produced, and observe the angle of elevation ADC. Let BD = a, L ABC = 0, 2 ADB ф. Then, Euc. I. 32, 2 BAD = 0 – 0. AB sin ADB AB sin • Now, BD sin BAD sin (0 - 0)' :: AB = (1). sin (0-0) AC sin ABC sin e, .. AC = AB sin e, D or a sin φ a. Again, AB sin o sin φ or, from (1) AC sin (0 - 0) Ex. Let BD 120, 0 = 60°, $ 45°. sin 60° sin 45° Then, A0 = 120. sin (60° - 45°) = 120 1 120, y V3 1 sin 60° sin 45° 2 12 120. sin 15° 3 1 212 J3 = 60 (3 + 13). 3 1 46. To find the height of an inaccessible object when it is not convenient to measure any distance in a line with the base of the object. Let a distance BD be measured in any direction in the same horizontal plane as BC, and let the angles ABC, ABD, ADB be observed. Let BD a, LABC B, ZADB sin or {180° (B + y)}' AB sin .....(1). sin (8 + x)' sin (8 + 9) sin or from (1), AC D sin (8 + x) 47. To find the distance of an object by observation from the top of a tower whose height is known. Let B be the object in the same B horizontal plane with c the foot of the tower, and let the angle of depression DAB be observed. Let AC = h, 2 DAB Now, BD sin sin y .: AB = Ꮂ . a sin' a, :: AC AB.sina, sing = a. .: BC BC BC Then we have, = cot ABC = cot DAB, or = cot , AC h cot 0, the distance required. CoR. Suppose B to be not in the same level with C. Let m be the height of B above C, then B is on the same level with a point which is h m from A. i. BC = (h – m) cot 0. Ex. IX. 1. Find the height of a tower 200 yards distant when it subtends an angle of 15°. 2. From the top of a tower, the angle of depression of a point in the horizontal plane at the foot of the tower was 30°. Given the height to be 60 ft., find the distance of the point. 3. The angle of depression of two consecutive milestones in a direct line with the summit of a hill were observed to be 60° and 30°. Find the height of the hill. 4. An object is observed from a ship to be due E. After sailing due S. for six miles it is observed to be N.E. Find the distance from the last position of the ship. 5. There is an object A, and two stations B and C are taken in the same plane. Given that BC 50, 2 ABC 60°, L ACB = 30°, find AB, AC. 6. The elevation of a tower is found to be 45°, and on approaching 60 feet nearer the elevation is 75o. Find the height of the tower. 7. Wishing to know the breadth of a river I observed an object on the opposite bank, and, having walked along the side of the river a distance of 100 yards, found the angle subtended by the object and my first station to be 30°. Find the breadth of the river. 8. A person, standing exactly opposite to the centre of an oblong which measures 16 ft. by 12 ft., and such that the line drawn from the centre to his eye is at right angles to the oblong, observes that the diagonal subtends an angle of 60°. Find his distance. 9. The angles of elevation of the summit of one tower, whose height is h, are observed from the base and summit of |