+14917 Find the inradius of triangle JKL if JK=JL=17 and KL=18.
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\(18^2=17^2+17^2-2\cdot17^2\cdot cos\ J\\ cos \ J=\dfrac{2\cdot 17^2-18^2}{2\cdot17^2}=0.439446\\ \color{blue}J=63.931°\\ \color{blue}K=L=58.034°\)
\(y=tan\dfrac{J}{2}\cdot x=-tan\dfrac{K}{2}\cdot (x-17)\\ tan\dfrac{J}{2}\cdot x+tan\dfrac{K}{2}\cdot x=tan\dfrac{K}{2}\cdot 17\\ x= \dfrac{tan\dfrac{K}{2}\cdot 17}{tan\dfrac{J}{2}+tan\dfrac{K}{2}}=\dfrac{tan\dfrac{58.034°}{2}\cdot 17}{tan\dfrac{63.931°}{2}+tan\dfrac{58.034°}{2}}\)
\(x=8\) \(y=tan\dfrac{J}{2}\cdot x=tan\dfrac{63.931°}{2}\cdot x\)
\(y=4.992\)
The inradius of triangle JKL is 4.992.
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