Question

Find the measure of an angle which is ${36}^{\circ }$ more than its complement.

Answer 1

Let $x$ be the one angle. So, the complement of $x={90}^o-x$.

Given, an angle is ${36}^o$ more than its complement.

$\Rightarrow$ $x=({90}^o-x)+{36}^o$

$\Rightarrow$ $x={90}^o-x+{36}^o$

$\Rightarrow$ $x+x={126}^o$

$\Rightarrow$ $2x={126}^o$

$\Rightarrow x=\frac{{126}^o}{2}$

$\Rightarrow x={63}^o$.

Therefore, complementary of $x$,

$={90}^o-x$

$={90}^o-{63}^o$

$={27}^o$.

Hence, the required angle is $x={63}^o$.