Question

Find the complement angle of:

(vi). $\frac{1}{6}$ of ${90}^{\circ }$

Answer 1

Step: Finding the complementary angle

We know that, two angles are said to be complementary when their sum is ${90}^{\circ }$

Let complimentary angle of $\frac{1}{6}$ of ${90}^{\circ }$ be $x^{\circ }$

$\therefore x^{\circ }+\frac{1}{6}$ of ${90}^{\circ }={90}^{\circ }$

$\Rightarrow x^{\circ }+\frac{1}{6}\times {90}^{\circ }={90}^{\circ }$

[BODMAS rule]
$\Rightarrow x^{\circ }+{15}^{\circ }={90}^{\circ }$

$\therefore x^{\circ }={75}^{\circ }$

Hence, complementary angle of $\frac{1}{6}$ of ${90}^{\circ }$ is ${75}^{\circ }$