+1927 In triangle ABC, \cos \angle A = \frac{1}{8}. Find the length of angle bisector \overline{AD}.
+130060 Assuming a right triangle ????
AB = sqrt [ 8^2 -1^2] = sqrt (63)
Let x = CD 8 - x = BD
Angle BAC = 90
Angle DAC = 45
sin DAC = sqrt (2) / 2
sin BCA = sqrt (63) / 8
Because AD is a bisector, we have that
CA/CD = AB/ BD
1/x = sqrt (63)/(8-x)
8 - x = x sqrt (63)
x + xsqrt (63) = 8
x ( 1 + sqrt 63) = 8
x = 8 / ( 1 + sqrt 63) = CD
Law of Sines
CD / sin CAD = DA / sin DCA
[8/ (1 + sqrt 63)] / (sqrt (2) / 2 ] = DA / [sqrt (63) / 8]
DA = [ sqrt (63) / 8 ] [ 8 / [ 1 + sqrt (63)' / (sqrt (2) /2)
DA = [sqrt (63) / (1 + sqrt 63) / (1/sqrt 2]
DA = sqrt (2) * sqrt (63) / (1 + sqrt 63 ) ≈ 1.26 = length of angle busector
