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Let AB be a diameter of a circle, and let C be a point on the circle such that AC=8 and BC=4 The angle bisector of ACB intersects the circle at point M Find CM.

 Feb 27, 2022
 #1
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7 times the square root of 2

 Feb 27, 2022
 #2
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One way -- I hope that it's not the easist way ...

 

I'm going to look at triangle(ACM) and find CM by using the Law of Cosines.

To do this, I need the length of AC (which is 8), the length of AM (which I have

to calculate), and the size of angle(A) (which I have to calculate).

 

First, note that since angle(ACB) is inscribed in a semicircle, it is a right angle.

this makes triangle(ACB) a right triangle.

 

To find the size of angle(A): 

   Looking at right triangle(ACB):  A  =  tan-1(CB/CA)  =  tan-1​(4/8) 

                                                     A  =  26.565o

 

To find the length of AM:

   First, find the length of AB:

        Pythagorean Theorem:  AB2  =  AC2 + BC  =  8+ 42  =  80

                                                AB  =  sqrt(80)  

 

   Now, use the Angle-Bisector Theorem to find the length of AM:

      AM / MB  =  AC / BC  =  8 / 4

 

   Let  AM  =  x     --->     This makes  MB  =  sqrt(80) - x

 

    AM / MB  =  x / [ sqrt(80) - x ]  =  8 / 4  =  2

    AM  =  x  =  2·[ sqrt(80 - x ]  =  2sqrt(80) - 2x

                x  =  2sqrt(80) / 3

 

 

Finally, use the Law of Cosines to find CM:

   CM2  =  AM2 + AC2 - 2 · AM ·AC · cos(A)

 

We can now finish this by replacement ...

 Feb 28, 2022

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