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# Help please due tommorow. INcredibly hard

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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 Jan 3, 2020
edited by Guest  Jan 3, 2020

#1
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By power of a point, CM works out to 8*sqrt(3).

Jan 3, 2020
#2
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I don't think it was the right answer

Jan 3, 2020
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can u post it in radical form

Jan 3, 2020
#5
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The exact radical form is 3*sqrt(13).

Jan 3, 2020
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Guest Jan 3, 2020
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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 Jan 4, 2020
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I added  Q in order to create the right triangle CMQ.

MQ is a diameter of a circle. MQ = 10

The angle CMQ = 8.13°                     cos(8.13°) = CM / 10

CM = 9.899

Dragan  Jan 4, 2020
edited by Dragan  Jan 4, 2020
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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

Jan 4, 2020
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See second post of same question here: https://web2.0calc.com/questions/help-meeee-due-tommorow-the-help-was-false

Jan 5, 2020