Q67 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

In $\triangle A B C, \frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3}+\frac{1+\cos B}{r_1+r_3}=$

  • A
    $\frac{2}{3 R}$
  • B
    $\frac{R}{2}$
  • Correct answer
    C
    $\frac{3}{2 R}$
  • D
    $\frac{6 R}{5}$

Solution

We know that, $\begin{aligned} r_1+r_2 & =4 R\left(\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}+\sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}\right) \\ & =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right] \\ & =4 R \cos \frac{C}{2} \sin \left(\frac{A+B}{2}\right)=4 R \cos \frac{C}{2} \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =4 R \cos \frac{C}{2} \cdot \cos \frac{C}{2}\end{aligned}$ $\begin{array}{cc} & =2 R\left(2 \cos ^2 \frac{C}{2}\right)=2 R(1+\cos C) \\ \therefore & \frac{1+\cos C}{r_1+r_2}=\frac{1}{2 R} \end{array}$ Similarly, $\begin{aligned} & \frac{1+\cos A}{r_2+r_3}=\frac{1}{2 R}, \frac{1+\cos B}{r_1+r_3}=\frac{1}{2 R} \\ \therefore \quad & \frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3}+\frac{1+\cos B}{r_1+r_3}=\frac{3}{2 R} \end{aligned}$

Hence, the correct answer is option: 'C'.

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