Question

In a $\Delta ABC$, 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,
$r_2+r_3=4R{\cos }^2\frac{A}{2}$
$\Rightarrow r_2+r_3=4R{\cos }^{2}45^{\circ }\left[\because \angle A=90^{\circ }\right]$
$\Rightarrow r_2+r_3=4R\times \frac{1}{2}$
$\Rightarrow r_2+r_3=2R$
$\therefore {\cos }^{-1}\left(\frac{R}{r_2+r_3}\right)={\cos }^{-1}\left(\frac{R}{2R}\right)={\cos }^{-1}\left(\frac{1}{2}\right)=60^{\circ }$