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In a $\triangle A B C$, if $\angle A=90^{\circ}$, then $\cos ^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to
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The correct answer is:
$60^{\circ}$
Given, $\angle \mathrm{A}=90^{\circ}$
In $\triangle A B C$, we know that
$\begin{aligned}
& r_2+r_3=4 \mathrm{R} \cos ^2 \frac{A}{2} \\
& \Rightarrow \quad r_2+r_3=4 \mathrm{R} \cos ^2 45^{\circ} \\
& \Rightarrow \quad r_2+r_3=4 \mathrm{R} \times \frac{1}{2} \\
& \Rightarrow \quad r_2+r_3=2 \mathrm{R} \\
& \therefore \quad \cos ^{-1}\left(\frac{R}{r_2+r_3}\right)=\cos ^{-1}\left(\frac{R}{2 R}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
\end{aligned}$
In $\triangle A B C$, we know that
$\begin{aligned}
& r_2+r_3=4 \mathrm{R} \cos ^2 \frac{A}{2} \\
& \Rightarrow \quad r_2+r_3=4 \mathrm{R} \cos ^2 45^{\circ} \\
& \Rightarrow \quad r_2+r_3=4 \mathrm{R} \times \frac{1}{2} \\
& \Rightarrow \quad r_2+r_3=2 \mathrm{R} \\
& \therefore \quad \cos ^{-1}\left(\frac{R}{r_2+r_3}\right)=\cos ^{-1}\left(\frac{R}{2 R}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
\end{aligned}$
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