CASE 2. Then the hypothenuse and a side are given. 'a = 10 + log a - log C......... ..(1), and B 90° A ..(2), also 72 co a> (c + a)(c a); ::. log b = } {log (c + a) + log (c – a)}..... (3). CASE 3. When an acute angle and a side are given. Let A, a be the given angle and side. Then B = 90° A.......... .(1), 6 also tan B, or log 6 = L tan B 10 + log a.....(2), a Also a C and= sin A, or log a – log log c = L sin A 10, or log c = 10 + logo L sin A. CASE 4. When the hypothenuse and an acute angle are given. Let A be the given acute angle. We have B 90° A...... ....(1). sin A, or log a = log c + L sin A - 10... (2). 6 And = cos A, or logo log c + L cos A - 10.... (3). It is evident, from Art. 30, that when the angles only of a triangle are known, we can determine the ratio only of the three sides of the triangle to each other. Ex. 1. Given A 23° 41', a = 35, solve the triangle. L tan B - 10 – log a log 35 10.3579092 10 + 1.5440680 1.9019772 log 79.795 ... 6 79.795. log 6 Also log c = 10 + log a L sin A 10 + log 35 - L sin 23° 411 log 87.134 10 + log a 9.8293545, 946. Also, by tables, D = 2724, 946 x 60". = 21" nearly. 2724 34° 1' 21". ..d or A Ex. VII. Log 32 1. Given a = 32, A = 63° 45', find b. 1:5051500, L cot 63° 45' = 9.6929750, Log 15780 = 4.1981070, log 15781 = 4.1901345. 2. Given c = 151, A = 37° 42', find a. Log 151 = 2.1789769, L sin 37° 421 = 9.7864157, Log 92340 = 4.9653899, tab. diff. = 47. 3. Given a = 60, c = 65, find b, A. Log 2 = 3010300, log 3 = 4771213, L sin 67° 23! 9.9652480. 4. Given a = 73, 6 84, find A, C. 2:0464479. 5. Given B = 71° 41' 10', c = 24, find b. Log 24 = 13802112, L cos 18° 18' = 9.9774609, Log 2278.5 = 3:3576490. Log 1044 = 3·0187005, log 458 = 2.6608655, Log 691.49 = 2.8397830. 75°, find a, b. 9. Given a = 17, c = 34, find A, b. 10. Given a = 5,6 = 5 73, find A. 11. Given a = 15°, find the length of the perpendicular from C on AB. 12. CD is the perpendicular from C on AB, and DE the perpendicular from Don BC. Given B 60°, a = 20, find DE 10, B 28, B CHAPTER IX. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three sides of a triangle, to find the remaining parts. 72 + 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 (8 - b) (8 -c) We have, sin 2 bc Α. 8(8 – a) 2 bc (s – b) (8 - c) 8(8 – a) COS 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. a sin B We have, Art. 30, 6 = sin A' a sin c sin A' both of which formula 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. b + c This formula is adapted to logarithmic computation, and determines } (B – C). We know also ! (B + C), for it is the complement of A. B are easily : b sin A Then, Art. 30, a which determines di sin B 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° - 4 A, and a = 28 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. (1.) Let the value of sin B be unity. b We then have sin A 1 sin 90°, and :: A Hence the triangle is right-angled at A. 90°. a (2.) Let the value of sin B be > 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 6—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, b, and L ABC = B; and in the triangle A'BC, we have BC = a, A'C = b, and L ABC = B. Again, the sides BA and BA correspond to the two values = a, AC |