56. If the base of a triangle be trisected, and k,, kz, kz be the cotangents of the angles which the parts subtend at the vertex, (kı + k) (k, + kg) = 4 (1 + k,). . 57. If a, b, c are in A. P., then also cos A, vers B, cos C are in A, P. 58. If a’, 6, ca are in A. P., then also cot A, cot B, cot C are in A. P. a + b 59. If are in A. P., then also b + c C+ a a с > cot A cot} A, cot B cot 1 B, cot C cot } C are in A. P. 60. If a2 + bc, b2 + ca, c + ab are in A.P., then also tan Ž A, tan 1 B, tan ; C are in A. P. 61. If a* + bc*, 64 + cʻa”, c4 +ab2 are in A. P., then also tan A, tan B, tan C are in A. P. 62. If AH be the bisector of A, AH”. (6+ c)2 = 4bcs (8 – a). 63. If x, y, z are the perpendiculars on the sides from any point within a triangle, ax + by + cz = - 28. 64. In the last proposition show that x2 + y2 + z2 is a minimum, when ./a=y/b=z/c=25/(a* + b2 +c). [In Examples (65—69), the letters a, b, c, A, B, C are not given as the elements of a triangle.] 65. Given that a = b cos C + c cos B, b= c cos A + a cos C, c = a cos B + b cos A, solve for cos A, cos B, cos C. 66. From the same equations, show that sin Asin B sin C _2 _{s (8 – a) (8 – b) (8 – c)}, b abc where 28 = a + b + c. 67. Given that A + B + C = 180°, b=c cos A + a cos C, c = a cos B + b cos, 4, . show that a = b cos C + c cos B. = a a с 68. Given that b sin A = a sin B, c= a cos B + b cos A, 2ab cos C = a2 + b2 – c?, show that one value of A + B + C is 180°, and that b sin C = c sin B, a= b cos C + c cos B. sin A sin B sin c 69. Given that and c= a cos B + b cos A, 6 find sin A, sin B, sin C in terms of a, b, c. 70. If D be the middle point of BC, and if AD and AD' make the same angle with the bisector of BAC, then BD' : D'C = ca : 62 and AD' : AD= 2bc : 62 + co. 71. If L be the foot of the perpendicular from A on BC, and if AL and AL' make the same angle with the bisector of BAC, then, (A, B, C being in descending order of magnitude,) AL BM CN A - B B-C A-C - 1. 2 2 72. If D and E be points in BC such that AD and AE make equal angles with the bisector of BAC, then BD. BE :CD.CE= c2 : . : + + 4 cos COS COS 73. If b+c:c+a: a + b 4:5 :6, then sin A : sin B : sin C = 7:5:3, cos A : cos B : cos C =-7:11:13 and A = 120°. = CHAPTER VII. SOLUTION OF TRIANGLES (WITHOUT LOGARITHMS]. 157. We have shown in Art. 29 that, when the values of three elements of a triangle including one length are given, the other elements of the triangle have a determinate value. The process of finding the unknown elements of a triangle, when a sufficient number of elements are given, is called the Solution of Triangles. 158. The solution is accomplished by means of the formulæ of the last chapter. Thus, from an equation involving four elements, the fourth element may be found when three of them are known. 159. We shall have to investigate in each case (1) whether or not the solution found represents any possible geometrical triangle, and (2) whether or not the solution gives more than one possible geometrical triangle. 160. The condition, algebraically considered, for a possible solution is that any quantity whose square root has to be taken must be positive. The conditions, geometrically considered, for a possible solution are that any length, any angle, and the ratio of any angle (except the cos, tan, sec, or cot of such angles as may be obtuse) must be represented in the solution by a positive value; and that the sine or cosine of any angle must be represented in the solution by a value numerically not greater than 1. Of course it will be assumed that the elements which are given are themselves possible. 161. There will be more than one solution, algebraically considered, wherever a square root has to be taken; for before this square root we may place a positive or negative sign. There will be more than one solution, geometrically considered, wherever an angle has to be found from its sine or cosecant; for such an angle may have either one of two values, supplementary to one another. Of course one or other of such alternative solutions must be rejected if it involves an impossible value for some other element. 162. The chief cases to be considered in the solution of triangles are those which correspond to the cases of exact equality of two triangles given in Art. 29. These we will now consider. CASE I. a .: b= sin A' с Given two angles and a side: as B, C, a. This determines for A a real, i.e. positive, value provided a sin B Since sin B sin A a sin c Since ..c= sin C sin A' sin A Since all the values involved are positive, b and c have each one real, i.e. positive, value. 164. Or, we may find b or c without first finding A. For a cosec C since cot B + cot C cosec C, .. b = 6 cot B + cot C If b, instead of a, had been given, the equations for finding a and c would be b sin C a = b sin ( (cot B+cot C), and c= sin B a This determines for A a real value, provided the above fraction is numerically not greater than 1. (1) Let the fraction be positive, i.e. b2 + > a?. > i.e. (assuming b not <c) a > b-c, i.e. a+c> b..............(2). Moreover, since b not <c, .. a fortiori, a + b > C..........(3); and, since b2 + co > a’, .. a fortiori, 5+C> A...... .(1). (2) Let the fraction be negative, i.e. a-> b2 +c. Then we must have 2bc > ao – 62 – 09, i.e. (6 + c)2 > a’, i.e. b+c> a ...... (1). Moreover, .:: a? - c>b2, but a-c<b, .. a+c>b.........(2), .: a? – b?>c, but a-b<C, :. a+b > c.........(3). Hence the condition that any one angle A may be real is that, of the sides given, any two must be together greater than the third. (Compare Euc. I. 20.) = ✓ { 166. Or, we may solve by means of the equation А (8 – 8 (a) This determines for tan : A a real value, provided that the fraction whose root has to be taken is positive : i.e. provided that two or none of the factors s Q, 8 – 6, 8 -c are negative. But two of the factors, such as s- a and 8 – b, cannot be negative, for then 28 - a - b, i.e. c would be negative. |