Question

Write the complement of ${51}^{\circ },{28}^{\circ },{13}^{\circ }$?

Answer 1

We know that, if two angles are complementary, then their sum is ${90}^o$.

Let $x$ be the one angle.

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

The complement of $x={51}^o$ is ${90}^o-x={90}^o-{51}^o={39}^o.$

The complement of $x={28}^o$ is ${90}^o-x={90}^o-{28}^o={62}^o.$

The complement of $x={13}^o$ is ${90}^o-x={90}^o-{13}^o={77}^o.$