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