Question

In the given figure AB is the side of a regular hexagon and BC is the side of a regular pentagon. Find $\angle$ OAC.

• A. ${24}^o$
• B. ${48}^o$
• C. ${20}^o$
• D. ${36}^o$

Answer 1

The correct option is A

${24}^o$

Given AB is the side of a regular hexagon.

$\Rightarrow$ $\angle AOB$ = $\frac{360}{6}={60}^o$

Given BC is the side of a regular pentagon.

$\Rightarrow$ $\angle BOC$ = $\frac{360}{5}={72}^o$

$\angle AOC$ = $\angle AOB$ + $\angle BOC$

= ${60}^o+{72}^o={132}^o$

OA = OC = radius of the circle.

$\therefore △COA$ is isosceles

$\Rightarrow \angle OAC=\angle OCA=\frac{{180}^o-{132}^o}{2}$

$\Rightarrow \angle OAC={24}^o$