Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side. 
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
  Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having length $a$ and $b$, what is the area of $ABC$?


An alternative approach: let we just find the inradius $r$. Assuming that $\widehat{C}=60^\circ$, $I_C$ is the projection of the incenter on the $AB$-side, $AI_C=a,BI_C=b$, we have:
$$\widehat{A}=2\arctan\frac{r}{a},\qquad \widehat{B}=2\arctan\frac{r}{b}\tag{1}$$
hence:
$$ \frac{\pi}{3} = \arctan\frac{r}{a}+\arctan\frac{r}{b}\tag{2} $$
must hold, from which:
$$ \sqrt{3} = \frac{\frac{r}{a}+\frac{r}{b}}{1-\frac{r^2}{ab}}\tag{3}$$
follows. Since:
$$ \Delta = \frac{1}{2}(a+\sqrt{3}\,r)(b+\sqrt{3}\,r)\sin 60^\circ =\frac{\sqrt{3}}{4}ab\left(1+\sqrt{3}\left(\frac{r}{a}+\frac{r}{b}\right)+3\frac{r^2}{ab}\right)\tag{4}$$
by exploiting $(3)$ we have:

$$ \Delta = \color{red}{\sqrt{3}\,ab},\tag{5}$$

pretty nice, don't you think?
