Question

In Fig. 10.11 AOC is a diameter of the circle and arc AXB = $\frac{1}{2}$ arc BYC Find $\angle BOC$

Answer 1

As arc $AXB=$ $\frac{1}{2}$ arc $BYC$

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

Also $\angle AOB+\angle BOC={180}^{\circ }$

Therefore $\frac{1}{2}\angle BOC+\angle BOC={180}^{\circ }$

or $\angle BOC=\frac{2}{3}\times {180}^{\circ }={120}^{\circ }$