Question

An angle is ${28}^o$ less than its complement, find its measure.

Answer 1

We know, two angles are said to be complementary, if the sum of their measures is ${90}^{\circ }.$

Let the angle be $x^{\circ }$ and its complement is ${90}^{\circ }-x^{\circ }$

Then, according to the statement,

$\Rightarrow x^{\circ }=({90}^{\circ }-x^{\circ })-{28}^{\circ }$

$\Rightarrow x^{\circ }+x^{\circ }={90}^o-{28}^{\circ }$

$\Rightarrow 2x^{\circ }={62}^{\circ }$

$\Rightarrow x^{\circ }=\frac{{62}^{\circ }}{2}={31}^{\circ }$.

Hence, the measure of the angle is $x={31}^{\circ }$

and its complement is ${90}^o-x={90}^o-{31}^o={59}^{\circ }$.