+1455 Let $\overline{AB}$ be a diameter of a circle, and let $C$ be a point on the circle such that $AC = 8$ and $BC = 4.5.$ The angle bisector of $\angle ACB$ intersects the circle at point $M.$ Find $\angle CMA.$
+129850 C
8 4.5
A B
M
Angle ACB = 90
AB = sqrt [ 8^2 + 4.5^2] = sqrt (84.25 ) ......radius = sqrt (84.25) / 2
Call the center of the circle O
OC = OA = sqrt (84.25) / 2
Law of Cosines
AC^2 = OC^2 + OA^2 - 2(OC * OA) cos (COA)
8^2 = 84.25 / 4 + 84.25 / 4 - 2 ( 84.25 / 4) cos (COA)
64 = 42.125 - 42.125cos (COA)
[ 64 - 42.125 ] / [-42.125 ] = cos (COA) = -175 / 337
arccos (-175/337) ≈ 121.28°
This is the measure of central angle COA
CMA is an inscribed angle intercepting the same arc as COA
So....its measure ≈ 121.28 / 2 ≈ 60.64°
