Question

Two complementary angles differ by ${10}^o$. 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 ${20}^o$.

$\Rightarrow$ $x+(x-{10}^o)={90}^o$

$\Rightarrow$ $2x-{10}^o={90}^o$

$\Rightarrow$ $2x={100}^o$

$\Rightarrow$ $x={50}^o$.

Then, the angles are $x={50}^o$ and other angle is $x-{10}^o={50}^o-{10}^o={40}^o$.