In a triangle ABC,if angle A = 90^{circ},then | vedclass

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(C) Given,$\angle A = 90^{\circ}$.In a $	riangle ABC$,we know the property $r_2 + r_3 = 4R \cos^2 \frac{A}{2}$.Substituting $\angle A = 90^{\circ}$:$

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Given,$\angle A = 90^{\circ}$.In a $	riangle ABC$,we know the property $r_2 + r_3 = 4R \cos^2 \frac{A}{2}$.Substituting $\angle A = 90^{\circ}$:$r_2 + r_3 = 4R \cos^2 \left(\frac{90^{\circ}}{2}ight) = 4R \cos^2 45^{\circ}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\cos^2 45^{\circ} = \frac{1}{2}$.Thus,$r_2 + r_3 = 4R 	imes \frac{1}{2} = 2R$.Now,calculating the expression:$\cos^{-1}\left(\frac{R}{r_2+r_3}ight) = \cos^{-1}\left(\frac{R}{2R}ight) = \cos^{-1}\left(\frac{1}{2}ight) = 60^{\circ}$.

In a $\triangle ABC$,if $\angle A = 90^{\circ}$,then $\cos^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to (in $^{\circ}$)

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