Question

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

• A. $\frac{7}{13}$
• B. $\frac{4}{3}$
• C. 13
• D. $\frac{13}{7}$

Answer 1

Answer: (C)

13

Answer 2

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