Question

If the given figure, AOC is a diameter of the circle and arc AXB = \[\frac{1}{2}\] arc BYC. Find ∠BOC.

 

Answer 1

We need to find  `angle BOC`

\[\text{ arc }  AXB = \frac{1}{2}\text{ arc}  BYC, \]

\[\angle AOB = \frac{1}{2}\angle BOC\]

\[\text{ Also }  \angle AOB + \angle BOC = 180°\] 

\[\text{ Therefore, } \frac{1}{2}\angle BOC + \angle BOC = 180° \]

\[ \Rightarrow \angle BOC = \frac{2}{3} \times 180°= 120°\]