Question

If $r:R:r_1=2:5:12$, then $A=$

• A. $45^{\circ }$
• B. $60^{\circ }$
• C. $30^{\circ }$
• D. $90^{\circ }$

Answer 1

Answer: (D)

We have
$r_1-r=4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$=4R\sin \frac{A}{2}\left(\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}\right)$
$=4R\sin \frac{A}{2}\cos \frac{B+C}{2}=4R{\sin }^2\frac{A}{2}$
Now, it is given that
$r:R:r_1=2:5:12$
$\Rightarrow 12k-2k=4(5k){\sin }^2\frac{A}{2}$
$\Rightarrow {\sin }^2\frac{A}{2}=\frac{1}{2}$
$\Rightarrow A=90^{\circ }$