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_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$r_2 = 4R \cos \frac{A}{2} \sin \f

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(C) We know that the exradii and inradius are given by:$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}$$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$Given $r_3 = r_1 + r_2 + r$,we have:$r_3 - r = r_1 + r_2$$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 $4R$ and factoring:$\sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}) = \cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2})$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 $A+B+C = \pi$,$\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$:$\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}$$	an^2 \frac{C}{2} = 1 \Rightarrow 	an \frac{C}{2} = 1$$\frac{C}{2} = \frac{\pi}{4} \Rightarrow C = \frac{\pi}{2}$Therefore,$A+B = \pi - C = \pi - \frac{\pi}{2} = \frac{\pi}{2} = 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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