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.
+26393 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
+26393 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
+1641 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.
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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
