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.

Guest Jul 15, 2021

#1**+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

heureka Jul 15, 2021

#1**+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

heureka Jul 15, 2021

#2**+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(4^{2} + 8^{2}) = √80

4/8 = √80/CQ CQ = (8√80)/4

PQ = sqrt(CP^{2} + CQ^{2})

**PQ = 18.02775638**

jugoslav Jul 15, 2021