Let triangle ABC have side lengths AB=13, AC=14, and BC=16. 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.
Compute the distance between the centers of these two circles.
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s=a+b+c2=16+14+132=21.5ρ=√(s−a)(s−b)(s−c)s=√(21.5−16)(21.5−14)(21.5−13)21.5=4.0383cosα=b2+c2−a22bc=142+132−1622⋅14⋅13α=arccos142+132−1622⋅14⋅13α=72.5754o
mf(x)=tanα2=tan72.57542=0.7342425
f(x)=0.7342425⋅x0.7342425⋅x1=ρ0.7342425⋅x1=4.0383x1=5.5
P1(5.5, 4.0383)
r=0.7342425⋅x2x2=rm(ρ+r)2=(r−ρ)2+(x2−x1)2(ρ+r)2=(r−ρ)2+(rm−x1)2(4.0383+r)2=(r−4.0383)2+(r0.7342425−5,5)2r=15.75 wolfram alpha
The distance between the centers of these two circles is
4.0383 + 15.75 = 19.788
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