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In the 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
 #1
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wHY IS NO ONE HELPING ME!

 Jan 3, 2020
 #2
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The answer is:   CM = 9.899497463    indecision

 Jan 4, 2020
 #3
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Can you explain the method used?

Guest Jan 4, 2020
 #4
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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    

 

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 indecision 

Dragan  Jan 4, 2020
 #5
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If this a competition question, assuming AB=MC is one way to get quick approx. 

so we know AB=10 ,thus MC=10 

well the real value of MC= 9.9

If it is MCQ, choose the value closer to 10 :D 

 Jan 4, 2020
edited by Guest  Jan 4, 2020
 #6
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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 units.

 Jan 4, 2020
 #7
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Where is the address of the original question!

 Jan 4, 2020
 #10
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See the original post here: https://web2.0calc.com/questions/help-please-due-tommorow-incredibly-hard

Guest Jan 5, 2020
 #11
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Thanks guest. :)

Melody  Jan 5, 2020
 #8
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Hi Melody: Sorry, I didn't realize that it had been posted twice! But, looking at the page, the original post is just 5 questions below this one. I just copied and pasted my answer under the original one.

 Jan 4, 2020
 #9
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Thanks,

Thanks for your answer too.

My comment was aimed at the asker.

 

Thje heading was "The help was false"

This meant that the asker was given and answer and was too impolite to continue with the question in the original thread.

Hence my request for the original thread address.

 

Could you please put a link on both so they are together, thanks.

As yours was not the only answer.

Melody  Jan 5, 2020

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