Question

An angle is $\frac{2}{3}$ of its complement. What is its magnitude?

Answer 1

We have, angle is $\frac{2}{3}$ of its complement.

Let the required angle be $x$.

Then, its complement will be $={90}^o-x$.

Given, that $x=\frac{2}{3}({90}^o-x)$

$⟹$ $3x=2\times ({90}^o-x)$

$⟹$ $3x={180}^o-2x$

$⟹$ $3x+2x={180}^o$

$⟹$ $5x={180}^o$

$⟹$ $x=\frac{{180}^o}{5}$

$⟹$ $x={36}^o$.

$\therefore$ Its complement $={90}^o-x$

$={90}^o-{36}^o$ $={54}^o$.

Hence, the required angle is ${36}^o$.