Question

Consider the figure given:
In the figure $O$ is the centre of the circle. $ABC$ is an isosceles triangle and $OBC$ is an equilateral triangle. Find $\angle A$ and $\angle ABO$.

Answer 1

Since $OBC$ is an equilateral triangle, $m\angle BOC={60}^o$
$\Rightarrow m\angle A={30}^o$ ...(angle subtended at the centre of a circle is double the size of the angle subtended at the edge from the same two points)
Now, $△ABC$ is an isosceles triangle,
Hence, $m\angle ABC=({180}^o-{30}^o)/2={75}^o$
$\Rightarrow m\angle ABO={75}^o-{60}^o={15}^o$