In a $\triangle \mathrm{ABC}$, if $r: \mathrm{R}: r_2=1: 3: 7$ then $\sin (\mathrm{A}+\mathrm{C})+\sin…

In a $\triangle \mathrm{ABC}$, if $r: \mathrm{R}: r_2=1: 3: 7$ then $\sin (\mathrm{A}+\mathrm{C})+\sin \mathrm{B}=$
  1. $0$
  2. $\sqrt{3}$
  3. $1$
  4. $2$

Solution

Let $r=\mathrm{k}, \mathrm{R}=3 \mathrm{k}$ and $r_2=7 \mathrm{k}$ $\begin{aligned} & r_2-r=4 \mathrm{R} \cos \frac{\mathrm{~A}}{2} \cdot \sin \frac{\mathrm{~B}}{2} \cdot \cos \frac{\mathrm{C}}{2}-4 \mathrm{R} \sin \frac{\mathrm{~A}}{2} \cdot \sin \frac{\mathrm{~B}}{2} \cdot \sin \frac{\mathrm{C}}{2} \\ & 7 \mathrm{k}-\mathrm{k}=4 \mathrm{R} \sin \frac{\mathrm{~B}}{2}\left[\cos \frac{\mathrm{~A}}{2} \cdot \cos \frac{\mathrm{C}}{2}-\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{C}}{2}\right] \\ & 6 \mathrm{k}=12 \mathrm{k} \sin \frac{\mathrm{~B}}{2} \cos \left(\frac{\mathrm{~A}+\mathrm{C}}{2}\right) \quad[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi] \\ & \frac{1}{2}=\sin \frac{\mathrm{B}}{2} \cdot \cos \left(\frac{\pi}{2}-\frac{\mathrm{B}}{2}\right)=\sin ^2 \frac{\mathrm{~B}}{2} \Rightarrow \sin \frac{\mathrm{~B}}{2}=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{~B}=90^{\circ} \end{aligned}$ $\begin{aligned} & \text { Now, } \sin (A+C)+\sin B=\sin (\pi-B)+\sin B \\ & =2 \sin B=2 \sin 90^{\circ}=2\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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