Question

In the given figure, $AB$ is a side of a regular hexagon and $AC$ is a side a regular eight sided polygon, Find:
$\angle OBC$

Answer 1

As $AB$ is the side of a hexagon so the
$\angle AOB=\frac{{360}^o}{6}={60}^o$
$AC$ is the side of an eight sided polygon so,
$\angle AOC=\frac{{360}^o}{8}={45}^o$
From the given figure we can see that :
$\angle BOC=\angle AOB+\angle AOC={60}^o+{45}^o={105}^o$
Again, from the figure we can see that $△BOC$ is an
isoceles triangle with sides $BO=OC$ as they are the radii of the same circle.
Angles $\angle OBC=\angle OCB$ as they are opposite angles to the equla sides of an isoceles triangle.
Sum of all the angles of a triangle is ${180}^o$
$\angle OBC+\angle OCB+\angle BOC={180}^o$
$2\angle OBC+\angle BOC={180}^o$
$2\angle OBC+{105}^O={180}^o$ as, $\angle OBC=\angle BOC$
$2\angle OBC={180}^o-{105}^o$
$2\angle OBC={75}^o$
$\angle OBC={37.5}^o={37}^o{30}^{\prime }$
As, $\angle OBC=\angle BOC$
$\angle OBC=\angle BOC={37.5}^o={37}^o{30}^{\prime }$