Q64 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

If $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$, then $x=$

  • A
    5
  • B
    7
  • Correct answer
    C
    13
  • D
    17

Solution

We have, $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$ $\Rightarrow \quad \sin ^{-1}\left(\frac{12}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{5}{x}\right)$ $\Rightarrow \quad \sin ^{-1}\left(\frac{12}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right)$ $\Rightarrow \quad \cos ^{-1}\left(\frac{\sqrt{x^2-144}}{x}\right)=\cos ^{-1}\left(\frac{5}{x}\right)$ $\Rightarrow \quad \frac{\sqrt{x^2-144}}{x}=\frac{5}{x}$ $\Rightarrow \quad \sqrt{x^2-144}=5 \Rightarrow x^2-144=25$ $\Rightarrow \quad x^2=144+25$ $\Rightarrow \quad x^2=169 \Rightarrow x=13$

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

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