Question

Two complementary angles differ by ${12}^o$, find the angles?

Answer 1

Let $x$ be the one angle.

So, the complement of $x={90}^o-x$.

Given, two complementary angles differ by ${20}^o$.

$\Rightarrow x-({90}^o-x)={12}^o$

$\Rightarrow x-{90}^o+x={12}^o$

$\Rightarrow 2x={12}^o+{90}^o$

$\Rightarrow x=\frac{{102}^o}{2}$

$\Rightarrow x={51}^o$.

Therefore, complementary of $x$,

$={90}^o-x$

$={90}^o-{51}^o$

$={39}^o$.

Hence, ${51}^o$ and ${39}^o$ are the complementary angles.