Question

In the figure, AOC is a diameter of the circle and arc $AXB=\frac{1}{2}$ arc BYC. Find $\angle BOC$.

Answer 1

In the figure, AOC is diameter arc AXB = $\frac{1}{2}$ arc BYC

$\therefore \angle AOB=\frac{1}{2}\angle BOC$

$\Rightarrow \angle BOC=2\angle AOB$

But $\angle AOB+\angle BOC={180}^o$

$\Rightarrow \angle AOB+2\angle AOB={180}^o$

$\Rightarrow 3\angle AOB={180}^o$

$\therefore \angle AOB=\frac{{180}^o}{3}={60}^o$

$\therefore \angle BOC=2\times {60}^o={120}^o$