Whence cos2 a = s (8 − a) (13) Dividing (11) by this equation, and writing the corresponding equations for the other angles, we find (8 — b) (8 — c); tan2 } α = tan'ẞ= 8 (8-a) By means of these equations we may compute two of the angles, and find the third by subtracting their sum from 180°. But in practice it is better to compute the three angles independently, and check the accuracy of the work by taking their sum. If this sum comes out materially different from 180°, there is some mistake in the work; if not, it may be presumed correct. Example. Given a 273.960, b = 198.632, c = 236.914; find the angles. log (8a), 1.907 37 log (86), 2.193 46 log (8 - c), 2.071 29 sum of logs, 6.172 12 log 8, 2.549 92 log H3, 3.622 20 log H, 1.811 10 log tan a, 9.903 73 log tan 6, 9.617 64 log tany, 9.739 81 The discrepancy of 0'.2 is the result of the unavoidable errors from the omission of the decimals of the logarithms beyond the fifth. Another check on the accuracy of the work is obtained by computing the modulus of the triangle from its three separate expressions (§ 56, 3), and noting whether they agree, thus: The three results agree within the unavoidable limits of error. 4. The base of a parallelogram is 13, each side is 6, and its lesser diagonal is 12. Find its angles. 5. If the sides of a parallelogram are a and b, and one diagonal is p, express its angles. 6. The parallel sides of a trapezoid are 12 and 17, and the nonparallel sides 6 and 7. Find its four angles. Suggestion. Divide the trapezoid into a triangle and a parallelogram. 7. In a triangle are given the two sides, p and q, and the medial liner from the vertex to the middle point of the base. Prove Base = √2p2 + 2q2 — 4r2. 8. Given the three medial lines r, r', r" of a triangle; find the sides of the triangle from the preceding result. Ans. 12+2p12 — r12; } √/2x2 + 2p”2 — r22; † √2r22 + 2′′2 — ri2. } √2r'+2r”3 — p3. 9. Of the bisectors of the angles at the base of a triangle, the one cuts the opposite side in the ratio m:n, and the other in the ratio pq. By means of the equation (10) express the cosine of the angle at the vertex of the triangle. 60. CASE IV. Given two sides and the included angle. This case may be readily solved by Theorem III., because if the given sides are b and c, and the given included angle is a, we have for the third side a = √/b2 + c2 — 2bc cos a. Then, having the three sides, the remaining angles may be found as in the last section. But there is a more convenient method founded on the following theorem : THEOREM IV. As the sum of any two sides is to their difference, so is the tangent of half the sum of the angles opposite these sides to the tangent of half their difference. Proof. From the equation bcbc sin 6+ sin y: sin 6 sin y. bc sin ẞ: sin y, we have, by composition and division, (Th. I.) (§ 44) (6+y) cos †(ß − y); (6+) sin (6 — y). expressing the proportion as a b + c sin (6+y) cos (67) (833) b+c:b-c: tan (6+y) : tan (6 − y). Q.E.D. (16) The solution is now obtained as follows: We have Because the angle a is given, the only unknown term of the proportion is tan 6 y). This is given by the equation By this equation we obtain ( − y), which being added to and subtracted from (6+) gives ẞ and y. The remaining side of the triangle may then be found by Case I. But when this side as well as the angles are required, a more elegant method may be followed, which will be explained in the next section. Example. Given b= 4.567, c = 3.456, a 56° 7'.8; find the remaining parts. 2 sin 61. If in the present case all three remaining parts are wanted, formulæ may be derived as follows: Adding these equations we have b + c sin ẞ+ sin y = α sin a = (a) (844) (18) sin (6-1) From the equations (c) and (b) we obtain ta, a sin (6 − y) = (b−c) cos fα, { a cos (ẞ - y) = (b+c) sin a;) which equations are readily solved by Prob. I. Chap. V. By taking the quotient of these equations we may readily deduce the relation (16). EXERCISES. 1. Given b=2956.2, c=9090.8, a=98° 29'.6; find ẞ, y, and a. 2. A surveyor lays off two lines from the same point: the one due north, 279.25 metres, the other east 15° north, 109.262 metres. How far apart are the ends of the lines, and what is the direction of the line joining them? 3. The sides of a parallelogram are 26 and 15, and one angle is 126° 52′.2. Find the lengths of the two diagonals and the angles which they make with the sides. 4. Given the two diagonals d and d' of a parallelogram and the angle & which they form; express the sides and angles of the parallelogram algebraically and compute them for the special case d = 5, d' = 6, 49° 18'. |