Question

In a $\Delta A B C$, if $\angle A=90^{\circ}$, then $\cos ^{-1}\left(\frac{R}{r_{2}+r_{3}}\right)$ is equal to

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

Answer 1

Answer: (C)

We know that,
$r_{2}+r_{3}=4 R \cos ^{2} \frac{A}{2}$
$\Rightarrow r_{2}+r_{3}=4 R \cos ^{2} 45^{\circ} \,\,\,\left[\because \angle A=90^{\circ}\right]$
$\Rightarrow r_{2}+r_{3}=4 R \times \frac{1}{2}$
$\Rightarrow r_{2}+r_{3}=2 R$
$\therefore \cos ^{-1}\left(\frac{R}{r_{2}+r_{3}}\right)=\cos ^{-1}\left(\frac{R}{2 R}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}$