In triangle ABC,if 2R + r = r_2,then angle B | vedclass

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(D) We are given the relation $2R + r = r_2$.Using the standard formulas for circumradius $R$,inradius $r$,and exradius $r_2$:$r_2 - r = 4R \sin \frac

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$\frac{\pi}{2}$

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(D) We are given the relation $2R + r = r_2$.Using the standard formulas for circumradius $R$,inradius $r$,and exradius $r_2$:$r_2 - r = 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$r_2 - r = 4R \sin \frac{B}{2} [\cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2}]$$r_2 - r = 4R \sin \frac{B}{2} \cos(\frac{A+C}{2})$Since $A+B+C = \pi$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cos(\frac{A+C}{2}) = \sin \frac{B}{2}$.Thus,$r_2 - r = 4R \sin^2 \frac{B}{2}$.Given $2R = r_2 - r$,we have $2R = 4R \sin^2 \frac{B}{2}$.Dividing by $2R$,we get $1 = 2 \sin^2 \frac{B}{2}$,which implies $\sin^2 \frac{B}{2} = \frac{1}{2}$.Therefore,$\sin \frac{B}{2} = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$.This gives $\frac{B}{2} = \frac{\pi}{4}$,so $B = \frac{\pi}{2}$.

In $\triangle ABC$,if $2R + r = r_2$,then $\angle B$ is equal to

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