Using side length of triangle, find radius of touching circle Question: 

The side lengths of a triangle are equal to lengths $8, 9$ and $10$. Find the exact value of the radius of the circle passing through the endpoints of the longest side and the midpoint of the shortest side. 

I am very confused as to where to begin to solve the problem. If I solve the angles of the triangle, how will that help me to find the radius of the circles?
 A: Let the triangle be $\triangle ABC$ with $AB=8$, $AC=9$, $BC = 10$, and let the midpoint of $AB$ be $M$. Then, our goal is to find the radius of the excircle of $\triangle BCM$. By the cosine law, we know that $$\cos\angle B=\frac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\frac{83}{160}$$
and similarly by the cosine law, $$\cos \angle B = \frac{BM^2+BC^2-MC^2}{2\cdot BM\cdot BC}=\frac{116-MC^2}{80}\implies MC = \sqrt{\frac{149}{2}}$$
Hence, by the sine law, the radius of the excircle of $\triangle BCM$ is $$R=\frac{MC}{2\sin \angle B} = \sqrt{\frac{149}{2}} \cdot \frac{80}{\sqrt{18711}}$$
A: Another proof of this that differs only slightly. I haven't seen the proof for the radius of an ex-circle, so while @Lazy Lee's method was fine, the last step just came out of the blue for me. Continuing from @Lazy Lee's answer, $\cos{\angle B}=\frac{83}{160}$ and $E=(MC)^2=\frac{149}{2}$. Let the ex-circle of $\triangle BCM = O$. It follows that $BO=OC=MO=r$. Then $\angle MOC = 2\angle B$ because the legs of $\triangle MOC$ and $\triangle MBC$ intercept the same arc $\newcommand{\tarc}{\mbox{$\frown$}} \newcommand{\arc}[1]{\stackrel{\tarc}{#1}} \arc{MC}$. Using the law of cosines on $\triangle MOC$ to find $r$, we see that $E=(MC)^2=2r^2(1-\cos{\angle 2B})$. Using the fundamental theorem of trigonometry, $\sin^2{\angle B}+\cos^2{\angle B}=1$, we see that $\sin^2{\angle B}= 1-\big(\frac{83}{160}\big)^2$. Then substituting $\cos{\angle 2B}$ with $\cos^2{\angle B}-\sin^2{\angle B}$, we arrive at the same conclusion when solving for r.
