When the angle A is obtuse, the point D will fall on the other side of A, and we have by Geometry (Prop. XIII. Bk. IV.), a2 = b2 + c2 + 2 b A D. B a But since BAD is now the supplement of DAC BAC, by Art. 77 we have AD c cos BAD=c cos BAC: Substituting this value of AD, we have as before, a2 = b2 + c2 2 bc cos A. c cos A. When A is a right angle and a the hypothenuse, cos▲ is zero (30), and (96) becomes a2 = b2 + c2, and thus the formula (96) is true, whatever the angle A may be. In like manner we have 114. The cosine of any angle of a plane triangle is equal to the fraction whose numerator is the sum of the squares of the containing sides, diminished by the square of the opposite side, and whose denominator is twice the product of the containing sides. For, by (96), a2 = b2 + c2 2 bc cos A, whence, Similarly, from (97) and (98), we have (99) 115. By these formulæ the angles of a triangle can be found when the sides are given, but they cannot be conveniently applied in computation by logarithms. We then subtract both members of formula (99) from 1, and obtain 2bc and, substituting for 1 cos A its value, 2 sin2 A, by (80), we have Let now 2 s = a+b+c, so that s is half the sum of the sides of the triangle; then a b c = 2 (sc), ab+c=2 (s—b). Substituting these values in the preceding equation, and reducing, we have The sine of half of any angle in a plane triangle is equal to the square root of half the sum of the three sides less one of the adjacent sides, into half the sum less the other adjacent side, divided by the rectangle of the two adjacent sides. 116. If 1 be added to both sides of (99), then, substituting for 1cos A its value, 2 cos2 4, by (81), we have 4 b c (105) Let now s half the sum of the sides of the triangle, as in Art. 115; then, b+c+ a = 2 s, b+c=a=2 (s—a). Substituting these values in the preceding equation, we have The cosine of half of any angle of a plane triangle is equal to the square root of half the sum of the three sides, into half the sum less the side opposite the angle, divided by the rectangle of the two adjacent sides. 117. Dividing (102) by (106), (103) by (107), and (104) by (108), we have, by (13), The tangent of half of any angle of a plane triangle is equal to the square root of half the sum of the three sides, less one of the adjacent sides, into half the sum less the other adjacent side, divided by half the sum, into half the sum less the side opposite the angle. SOLUTION OF RIGHT-ANGLED TRIANGLES. 118. In a right-angled triangle, the side opposite to the right angle is called the hypothenuse; that adjacent to the right angle, and upon which the triangle is supposed to stand, is called the base; and the other side adjacent to the right angle, the perpendicular. The base and perpendicular have been termed the sides about the right angle. Of the acute angles, that adjacent to the base has been termed the angle at the base, and the other the angle at the perpendicular. B Thus, let A B C be any right-angled triangle, with the right angle at C, then h represents the hypothenuse, b the base, p the perpendicular, A the acute angle at the base, and B the acute angle at the perpendicular. 119. In order to solve the triangle, two A elements other than the right angle must be given, one of them being a side. Hence there will be four cases in which there may be given, respectively, I. The hypothenuse and an acute angle. II. A side about the right angle and an acute angle. CASE I. 120. Given the hypothenuse and an acute angle. Let there be given, in the right-angled triangle A B C, the hypothenuse h and the acute angle A; to find the angle B, the perpendicular p, and the base b. To find B. The angle B is the comple- A ment of A (Art. 44); hence,. To find p and b. By (85) and (86) we have p = h sin A = h cos B, B C bh cos Ah sin B; or, by logarithms, log p = log hlog sin A= log h+log cos B, (112) log blog h+log cos A= log h+ log sin B. (113) That is, The logarithm of either side about the right angle is equal to the logarithm of the hypothenuse, plus the logarithmic sine of the opposite angle, or plus the logarithmic cosine of the adjacent angle. NOTE 1. As the logarithmic sine and cosine are increased by 10 (Art. 99), the resulting logarithm will be so much too great, and must be diminished by 10. This increase by 10 will affect the work wherever the logarithms of trigonometric functions are used. NOTE 2. The last figure of an answer may occasionally be found to differ from the one given in this work, when it has been obtained by the use of different formulæ or tables. The results are not, however, generally carried so far as to admit of such a difference. When two methods of solving give different results, that is inserted which is most accurate, whether obtained by the usual method or not. EXAMPLES. 1. Given the hypothenuse of a right-angled triangle equal to 1785.395 feet, and the angle at the base equal to 59° 37′ 42′′; to solve the triangle. Solution. The angle at the perpendicular 90° 59° 37′42′′ =30°22′18′′. Let, now, h=1785.395 feet and A— 59° 37′ 42′′, and we have, by (112) and (113), Ans. Angle at the perpendicular, 30° 22′ 18′′; perpendicular, 1540.37 feet; base, 902.708 feet. 2. Given the hypothenuse of a right-angled triangle equal to 25 yards, and one of the acute angles equal to 54° 30′; to solve the triangle. 3. Given the hypothenuse of a right-angled triangle equal to 173.2 feet, and one of the acute angles equal to 37° 2′ 43′′; required the other parts. Ans. Angle, 52° 57' 17"; sides, 104.34 feet and 138.24 feet. |