Question

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

Answer 1

Given AB is the side of a regular nonagon.

$\Rightarrow$ $\angle AOB$ = $\frac{360}{9}={40}^o$ (1 mark)

Given BC is the side of a regular decagon.

$\Rightarrow$ $\angle BOC$ = $\frac{360}{10}={36}^o$ (1 mark)

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

= ${40}^o+{36}^o={76}^o$

OA = OC = radius of the circle.

$\therefore △COA$ is isosceles

$\Rightarrow \angle OAC=\angle OCA=\frac{{180}^o-{76}^o}{2}$

$\Rightarrow \angle OAC={52}^o$ (1 mark)