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Let triangle ABC have side lengths AB=13, AC=14, and BC=15. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.

 Jul 15, 2021

Best Answer 

 #1
avatar+26172 
+2

Let triangle ABC have side lengths AB=13, AC=14, and BC=15.
There are two circles located inside angle BAC
which are tangent to rays AB, AC, and segment BC.
Compute the distance between the centers of these two circles.

 

My answer see: https://web2.0calc.com/questions/triangle-abc-has-ab-13-ac-14-bc-15-2-circles-in-angle

 

laugh

 Jul 15, 2021
edited by heureka  Jul 15, 2021
edited by heureka  Jul 15, 2021
 #1
avatar+26172 
+2
Best Answer

Let triangle ABC have side lengths AB=13, AC=14, and BC=15.
There are two circles located inside angle BAC
which are tangent to rays AB, AC, and segment BC.
Compute the distance between the centers of these two circles.

 

My answer see: https://web2.0calc.com/questions/triangle-abc-has-ab-13-ac-14-bc-15-2-circles-in-angle

 

laugh

heureka Jul 15, 2021
edited by heureka  Jul 15, 2021
edited by heureka  Jul 15, 2021
 #2
avatar+1442 
+3

Let triangle ABC have side lengths AB=13, AC=14, and BC=15. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 13     AC = 14     BC = 15

s = 1/2(13+14+15) = 21

A = sqrt(21(21-13()21-14)(21-15) = 84

r = A/s = 4

Using the law of cosines, we find that ∠A = 67.380135º

∠PAY = 33.6900675º

AZ = AY = r / tan(A/2) = 6            CW = CY = 8              BW = BZ = 7

CP = sqrt(42 + 82) = √80

4/8 = √80/CQ         CQ = (8√80)/4

PQ = sqrt(CP2 + CQ2)

PQ = 18.02775638

 

 Jul 15, 2021
edited by jugoslav  Jul 15, 2021
 #3
avatar+1442 
+1

I made an error! cheeky

 

4/8 = √80/CQ         CQ = (8√80)/4      Wrong!!!

 

4/√80 = 7/CQ          CQ = (7√80) / 4      Correct!!!

jugoslav  Jul 15, 2021

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