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 $si{n}^{-1}\frac{5}{x}=A$
$\therefore \frac{5}{x}=sinA$
$si{n}^{-1}\frac{12}{x}=B$
$\therefore \frac{12}{x}=sinB$
$\therefore A+B=\frac{\pi }{2}$
$\Rightarrow sinA=sin\left(\frac{\pi }{2}-B\right)=cosB$
$\sqrt{1-si{n}^{2}B}$
$\Rightarrow \frac{5}{x}=\sqrt{1-\frac{144}{x^2}}$
$\Rightarrow \frac{25}{x^2}=1-\frac{144}{x^2}$
$\Rightarrow \frac{169}{x^2}=1$
$\Rightarrow x^2=169$
$\Rightarrow x=13$