Question

Find the angle whose;

i) Complement is one sixth of its supplement.

ii) Supplement is four times its complement.

Answer 1

1)Let the angle be A degree
complement of an angle A is ${90}^{\circ }-A$
supplement of that angle A is ${180}^{\circ }-A$
Given the complement is $\frac{1}{6}$th of supplement.
$90-A=\frac{1}{6}\times (180-A)$
$=30-\frac{A}{6}$
$\Rightarrow A-\frac{A}{6}=90-30$
$\frac{5A}{6}={60}^{\circ }$
$A=60\times \frac{6}{5}$
$\therefore A={72}^{\circ }$
2)
Let one angle be B degree and its supplement be $180-B$ degrees.
And Its Complete be $90-B$ degrees.
Given condition: $180-x=4(90-x)$
$\Rightarrow 180-x=360-4x$
$\Rightarrow 4x-x=360-180$
$3x=180$
$\therefore x={60}^{\circ }$