Exercise 396

In $\triangle ABC$, $AB=13$, $BC=14$, and $CA=15$. What is $\tan\dfrac A2$?

Solution

The semiperimeter is $s=21$. By Heron's formula the area is $\sqrt{21\cdot7\cdot6\cdot8}=84$, so the inradius is $r=\frac{84}{21}=4$.
The incircle touches $AB$ at a point $T$ with $AT=s-BC=21-14=7$, and the incenter $I$ lies on the bisector of $\angle A$ with $IT\perp AB$, $IT=r=4$.
In right triangle $ATI$, $\tan\dfrac A2=\dfrac{IT}{AT}=\boxed{\dfrac47}$.