Question

${\sin }^{-1}\frac{5}{x}+{\sin }^{-1}\frac{12}{x}=\frac{\pi }{2},(x>0)$ then $x=$

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

Answer: (B)

$13$

We have,

${\sin }^{-1}\frac{5}{x}+{\sin }^{-1}\frac{12}{x}=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}\frac{5}{x}=\frac{\pi }{2}-{\sin }^{-1}\frac{12}{x}$

$\Rightarrow {\sin }^{-1}\frac{5}{x}={\cos }^{-1}\frac{12}{x}$

$\Rightarrow {\sin }^{-1}\frac{5}{x}={\sin }^{-1}\sqrt{1-{\left(\frac{12}{x}\right)}^2}$

$\Rightarrow {\sin }^{-1}\frac{5}{x}={\sin }^{-1}\sqrt{\frac{x^2-144}{x^2}}$

$\Rightarrow {\left(\frac{5}{x}\right)}^2=\frac{x^2-144}{x}$

$\Rightarrow \frac{25}{x^2}=\frac{x^2-144}{x}$

$\Rightarrow 25=x^2-144$

$\Rightarrow x^2=169$

$\Rightarrow x=13$

Hence, this is the answer.