Three times the complement of an angle is ${20}^o$ more than its supplement. Find the angle.

Let $x$ be the required angle.

According to the question-

$3(90°-x)=180°-x$

$270°-3x=180°-x$

$\Rightarrow 3x-x=270°-180°$

$\Rightarrow x=\frac{90°}{2}=45°$