If in a triangle ABC,r_3 = r_1 + r_2 + r,then | vedclass

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(C) We know that the exradii and inradius are given by:$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$r_1 = 4R \sin \frac{A}{2} \cos \fra

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(C) We know that the exradii and inradius are given by:$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$Given $r_3 = r_1 + r_2 + r$,we have:$r_3 - r = r_1 + r_2$$4R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \cos \frac{C}{2} \sin \frac{A}{2} \cos \frac{B}{2} + 4R \cos \frac{C}{2} \cos \frac{A}{2} \sin \frac{B}{2}$Dividing by $4R$:$\sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}) = \cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2})$Using the identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$:$\sin \frac{C}{2} \cos(\frac{A+B}{2}) = \cos \frac{C}{2} \sin(\frac{A+B}{2})$Since $A+B+C = \pi$,we have $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$:$\sin \frac{C}{2} \cos(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2} \sin(\frac{\pi}{2} - \frac{C}{2})$$\sin \frac{C}{2} \sin \frac{C}{2} = \cos \frac{C}{2} \cos \frac{C}{2}$$\sin^2 \frac{C}{2} = \cos^2 \frac{C}{2}$$	an^2 \frac{C}{2} = 1 \Rightarrow 	an \frac{C}{2} = 1$ (as $C$ is an angle of a triangle,$\frac{C}{2} > 0$)$\frac{C}{2} = 45^{\circ} \Rightarrow C = 90^{\circ}$Since $A+B+C = 180^{\circ}$,$A+B = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

If in a $\triangle ABC$,$r_3 = r_1 + r_2 + r$,then $\angle A + \angle B$ is equal to (in $^{\circ}$)

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