Solve arc $\sin \frac{5}{x}+\arcsin \frac{12}{x}=\frac{\pi }{2}$.

Method-1: 

Given that $\arcsin \frac{5}{x}+\arcsin \frac{12}{x}=\frac{\pi }{2}$

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

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

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

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

Squaring on both sides 

$\frac{25}{x^2}=1-\frac{144}{x^2}\Rightarrow \frac{25}{x^2}+\frac{144}{x^2}=1$

$\Rightarrow 169=x^2$

$\Rightarrow x=13$ $(\because x>0)$

Method-2: 

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

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

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

$\Rightarrow A+B=\frac{\pi }{2}\Rightarrow B=\frac{\pi }{2}-A$

$\Rightarrow \sin B=\cos A=\sqrt{1-{\sin }^{2}A}$

$\Rightarrow \frac{12}{x}=\sqrt{1-\frac{25}{x^2}}\Rightarrow \frac{144}{x^2}+\frac{25}{x^2}=1\Rightarrow x^2=169$

$\Rightarrow x=13$$(\because x>0)$