In right triangle, ABC, the length of side AC is 8 the length of side BC is6 and C is 90 degrees The circumcircle of triangle ABC is drawn. The angle bisector of ACB meets the circumcircle at point M Find the length CM

Guest Jan 3, 2020

edited by
Guest
Jan 3, 2020

#7**+1 **

AC = 8

BC = 6

AB = ? AB = sqrt (6² + 8²) AB = 10 (AB is a diameter of a circle)

∠ACM = 45°

∠BAC = ? tan(BAC) = 6/8 BAC = 36.87°

Let the intersection point of CM and AB be an N.

∠ANC = 98.13°

Connecting M with Q which is exactly on the opposite side of the circle.

Now we have a right triangle CMQ

∠CMQ = 8.13°

And finally... cos(CMQ) = CM / MQ CM = 9.899497463

Dragan Jan 4, 2020

#10**+1 **

**CPhill: Doesn't the "Intrsecting Chord Theorem" apply to this question? If yes, then the diameter AB =10, which is intersected by CM, say at point "O", has lengths of AO =5.714 and BO =4.286 and and CO=4.849. Let MO =x, then we have: 5.714 * 4.286 =x * 4.849 and x =5.051. Therefore, the length of CM =4.849 + 5.051 =9.9 unit**

Guest Jan 4, 2020