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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 $\angle CMA.$

 Mar 19, 2024
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                   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°

 

 

cool cool cool

 Mar 20, 2024
edited by CPhill  Mar 20, 2024
edited by CPhill  Mar 20, 2024

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