If in a $\triangle A B C, r_3=r_1+r_2+r$, then $\angle A+\angle B$ is equal to

If in a $\triangle A B C, r_3=r_1+r_2+r$, then $\angle A+\angle B$ is equal to
  1. $120^{\circ}$
  2. $100^{\circ}$
  3. $90^{\circ}$
  4. $80^{\circ}$

Solution

We know that, $ \begin{aligned} & r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ & r_1=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ & r_2=4 R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} \\ & r_3=4 R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \end{aligned} $ Given that, $ \begin{aligned} & r_3=r_1+r_2+r \\ & \Rightarrow \quad r_3-r=r_1+r_2 \\ & \Rightarrow \quad 4 R \sin \frac{C}{2}\left(\cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{B}{2} \sin \frac{A}{2}\right) \\ & =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2} \sin \frac{B}{2}\right] \\ & \Rightarrow \quad \sin \frac{C}{2}\left(\cos \left(\frac{A+B}{2}\right)\right)=\cos \frac{C}{2}\left(\sin \left(\frac{A+B}{2}\right)\right) \\ & \Rightarrow \sin \frac{C}{2}\left(\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)\right)=\cos \frac{C}{2}\left(\sin \left(\frac{\pi}{2}-\frac{C}{2}\right)\right) \\ & {\left[\because A+B+C=\pi \Rightarrow \frac{\mathrm{A}}{2}+\frac{B}{2}=\frac{\pi}{2}-C\right]} \\ & \Rightarrow \quad \sin ^2 \frac{A}{2}=\cos ^2 \frac{C}{2} \\ & \Rightarrow \quad \tan \frac{C}{2}=1 \\ & \end{aligned} $ $ \Rightarrow \quad \frac{C}{2}=\frac{\pi}{4} \Rightarrow C=\frac{\pi}{2} $ We know that, $A+B+C=\pi$ $ \begin{array}{ll} \Rightarrow & A+B=\pi-\frac{\pi}{2} \\ \Rightarrow & A+B=\frac{\pi}{2} \end{array} $

Asked in: AP EAMCET 2004

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