Question

How many degrees are there in an angle which is on-fifth of its complement?

Answer 1

Let n be the angle $\left({90}^{0-n}\right)$ is complement of$n$

$\frac{{90}^0-n}{5}$

(one fifth of the complement angle )

for given $n=\frac{({90}^0-n)}{5}$

$\Rightarrow$ $6n={90}^0$

$n=15$

hence ${90}^0-n={90}^0-15={75}^0$

therefore,

${15}^0$ and ${75}^0$ are angles.