Log 24 = 5. Given B 71° 41' 10', c = 24, find 6. 1:3802112, L cos 18° 18' = 9.9774609, L cos 18° 19' = 9.9774191, log 22784 = 3.3576300, Log 2278.5 = 3.3576490. 293, c = 751, find b. Log 691.49 = 2.8397830. find a, b. 9. Given a 17, c = 34, find A, b. 10. Given a = 5,6 = 5/3, find A. 3 11. Given a 28, B = 15°, find the length of the perpendicular from C on AB. 12. CD is the perpendicular from Con AB, and DE the perpendicular from Don BC. Given B = 60°, a = 20, find DE 10, B 75°, CHAPTER IX. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three sides of a triangle, to find the remaining parts. 62 + ca a? We have, Art. 31, cos A from which A 2 bc may be determined ; and from similar formulæ we may find B and C. These formulæ are not however adapted to logarithmic computation. We shall therefore find it generally advisable to use the formulæ of Art. 34. A We have, sin bc Α. 8 a) 2 bc ve 1 (s – b) (8 – c) tan 8(8 - a) (s 25 COS = a sin B C From either of these formulæ we can determine A, and from similar formulæ determine the other angles. 39. Given one side and two angles, to find the remaining parts. Of course the third angle is at once known. Let a be the given side. We have, Art. 30, 6 sin A sin A' both of which formulæ are adapted to logarithmic computation. 40. Given two sides and the included angle, to find the remaining parts. Let b, c be the given sides, and A the included angle. cot A. This formula is adapted to logarithmic computation, and determines } (B - C). We know also } (B + C), for it is the complement of } A. Hence, B and C are easily determined: b sin A Then, Art. 30, a = which determines di C b + c sin NOTE.—When the two given sides are equal, the solution may be effected more easily by drawing a perpendicular from the given angle upon the opposite side, and so bisecting it. By drawing a figure, it is easily seen that, in this case, B = C = 90° – i A, and a = 25 cos B. 41. Given two sides and an angle opposite to one of them, to find the remaining parts. Let a, b, B be the given elements. , a a (1.) Let the value of sin B be unity. 7 We then have sin A = 1 sin 90°, and :: A = 90°. Hence the triangle is right-angled at A. . a a (2.) Let the value of sin B be > 1. 1 6 We then have sin A > 1, which is impossible. Hence, in this case, it is impossible to form a triangle with the given elements. (3.) Let the value of sin B be < 1. 6 Then, since, Art. 15, the sine of an angle is the same as the sine of its supplement, there are two values of A which satisfy the equality, sin A = 7 sin B, and these values are supplementary Let A, A' be the two values, then the relation between them is A + A' = 180°. If a is not greater than b, then A is not greater than B, and there is no doubt as to which value of A is to be taken. If, however, a is greater than that is, if the given angle is opposite to the less of the given sides, we must have A greater than B, and both values of A may satisfy this condition. This particular case, when the given angle is opposite to the less of the given sides, is called the ambiguous case. We will illustrate this geometrically. 42. The ambiguous case. Draw the line BC equal to the given side a, and draw BA making an angle B with BC. Then with centre C and radius Each of the triangles ABC, For, in the triangle ABC, we have BC a, AC = b, and LABC B; and in the triangle A 'BC, we have BC a, AC b, and L ABC B. Again, the sides BA and BA correspond to the two value: с of c which are obtained from the two values of A in the a equality sin A = sin B (see last Art.). 7 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' ad c = 2 bcos A. C = 6, A = 60°. 2. a = 18,6 4. a = 12, B = Ex. VIII. Solve the following triangles, having given1. 6 12, c 18 V2, A = 30°. 3. a = 5 73,6 = 5 2,c = $(/6 + 12). 12,6 60°, C = 15°. Log 121 = 2.0827854, L cot 20° 16' = 10.4326795, 28° 24', find A. 2.3747483, L sin 28° 24' = 9•6772640, 2.5327544, L sin 19° 18' 9:5191904, Log 2 :3010300, L cot 13° 16' 10:6275008, 18, find A, B. • L tan 24° 5' = 9.6502809, L tan 24° 6' 9.6506199, L tan 29° 12' = 9.7473194, L tan 29° 13' = 9.7476160. 8. a = 237, = Log 341 10. a 14, 3,6 = 11. a = 2 A = 60°, find B, C, and c. • 7, find A. Log 3 = .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)2 tano A 4 a. 14. If a, b, A are given, show from the equation, 72 + ca 2 bc cos A = a*; that if c and c' be the two values of the third side cc' 72 a’, and c + c' 2 b 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 18. If in a triangle tan ! (B − C) = tan, cot , show that b cos q = c. с 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. |