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)

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$