Question

Find the angles, when:

(i)The angles are complementary and the smaller is ${40}^{\circ }$ less than the larger.

(ii)The angles are complementary and the larger is four times the smaller.

Answer 1

We know, two angles are said to be complementary, if the sum of their measures is ${90}^{\circ }.$

Then:

$i)$ If one angle is $x$, then its complement given is $x+{40}^{\circ }$.

Then, $x+x+{40}^{\circ }={90}^{\circ }$

$⟹$ $2x={90}^o-{40}^o={50}^o$

$⟹$ $x=\frac{{90}^o}{5}={18}^o$.

Then, smaller angle$x={25}^{\circ }$ and larger angle is $x+{40}^o={25}^o+{40}^o={65}^{\circ }.$

$ii)$ If one angle is $x$, then its complement given is $4x$.

Then, $x+4x={90}^{\circ }$

$⟹$ $5x={90}^o$

$⟹$ $x=\frac{{90}^o}{5}={18}^o$.

Then, smaller angle$x={18}^{\circ }$ and larger angle is $4x=4\times {18}^o={72}^o$.