Trigonometry! #23

Notation: $r$ is the inradius, $r_{3}$ is the exradius of the circle with respect to $C$ and $R$ is the circumradius.

We know that $r=4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$

$\Rightarrow \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=\frac{1}{10}$ ....................(1)

We also know that $r_{3}=4R\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$

$\Rightarrow \cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=\frac{3}{5}$ ..........................(2)

Subtracting (1) from (2), we get

$\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)-\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=\frac{3}{5}-\frac{1}{10}$

$\sin\left(\frac{C}{2}\right)\left(\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)-\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\right)=\frac{1}{2}$

$\sin\left(\frac{C}{2}\right)\cos\left(\frac{A+B}{2}\right)=\frac{1}{2}$

$2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=1$

$\sin(C)=1$

$\therefore\boxed{C=90^{\circ}}$

Let inradius,Circumradius, exradius(opposite to A),exradius(opposite to B) and exradius(opposite to C) be respectively r,R,r1,r2 and r3, We know that incase of right triangles r+r1+r2=r3.....................(1),another theorem for any triangle r1+r2+r3-r=4R or r1+r2+12-2=4 * 5 or r1+r2=10,now putting this to (1) we get 2+10=12 or 12=12,so it satisfies the equation,means C=90 degrees