Q68 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

In a triangle ABC , if $a=5, b=3, c=7$, then $\sqrt{\frac{\sin (A-B)}{\sin (A+B)}}=$

  • Correct answer
    A
    $\frac{4}{7}$
  • B
    $16$
  • C
    $36$
  • D
    $\frac{4}{5}$

Solution

$\begin{aligned} & \text { } \sin (A+B)=\sin C \\ & \cos B=\frac{25+49-9}{2 \times 5 \times 7}=\frac{13}{14} \\ & \cos A=\frac{9+49-25}{2 \times 3 \times 7}=\frac{33}{2 \times 3 \times 7}=\frac{11}{14} \\ & \frac{\sin (A-B)}{\sin (A+B)}=\frac{\sin A \cos B-\cos A \sin B}{\sin C} \\ & =\left(\frac{\sin A}{\sin C}\right)+\frac{13}{14}-\left(\frac{\sin B}{\sin C}\right) \cos A \\ & =\frac{a}{c} \times \frac{13}{14}-\frac{b}{c} \times \frac{11}{14}=\frac{5}{7} \times \frac{13}{14}-\frac{3}{7} \times \frac{11}{14}=\frac{16}{49} \\ & \Rightarrow \sqrt{\frac{\sin (A-B)}{\sin (A+B)}}=\sqrt{\frac{16}{49}}=\frac{4}{7} .\end{aligned}$

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

Relevant Questions