18. The sides of a triangle ABC are in arithmetical progres b√3 sion; show that its area = √(2 ab) (36 - 2a). CHAPTER VIII. SOLUTION OF RIGHT-ANGLED TRIANGLES. 37. A triangle can always be determined when any three elements, with the exception of the three angles, are given. In the latter case we have only the same data as when two angles are given, for the third can always be found by subtracting the sum of the other two from two right angles (Euc. I., 32). Hence a right-angled triangle can always be determined when any two elements, other than the two acute angles, are given besides the right angle. And when one of the acute angles is given, the other may be obtained by subtracting it from a right angle. We have the following cases: CASE 1. When the two sides containing the right angle are given. A We shall take C as the right angle Hence the three elements, A, B, c, are determined. .(2). .(3). CASE 2. When the hypothenuse and a side are given. .. log b = (c + a) (c − a); {log (c + a) + log (c a)}.....(3). CASE 3. When an acute angle and a side are given 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). = cos A, or log b = log c + Lcos A 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. 3. Given α = = 151, A = 37° 42', find a. = = = 9.6929750, = 2.1789769, L sin 37° 42/ 9-7864157, = 60, c = 65, find b, A. 3010300, log 3 4771213, = 1.8129134, Lsin 67° 22 9.9651953, Lsin 67° 23 = 9.9652480. 4. Given a = 73, b = 84, find A, c. Log 73 = 1.8633229, L tan 40° 59/ = 9.9389079, 9.9391631, Log 84 1.9242793, L tan 41° = = Lsin 40° 59 = 9.8167975, L sin 41° = 9·8169429, Log 111.288 = 2.0464479. = = Lcos 18° 19' 9.9774191, log 2278.4 = Log 2278.5 3.3576300, Log 1044 3.0187005, log 458 = 2-6608655, = 2.8397830. 7. Given a 8. Given c = 9. Given a = 12. CD is the perpendicular from C on AB, and DE the perpendicular from D on BC. Given B = 60°, a = 20, find DE. CHAPTER IX. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three sides of a triangle, to find the remaining parts. We have, Art. 31, cos A = b2 + c2 a2 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. 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, b a sin B C = 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. determines (B We know also C). (B+ C), for it is the complement of A. Hence, B and C are easily determined. Then, Art. 30, a = b sin A which determines a. 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° – A, and a = 2b 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. |