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Question
If the given figure, AOC is a diameter of the circle and arc AXB = \[\frac{1}{2}\] arc BYC. Find ∠BOC.
Solution
We need to find `angle BOC`
\[\text{ arc } AXB = \frac{1}{2}\text{ arc} BYC, \]
\[\angle AOB = \frac{1}{2}\angle BOC\]
\[\text{ Also } \angle AOB + \angle BOC = 180°\]
\[\text{ Therefore, } \frac{1}{2}\angle BOC + \angle BOC = 180° \]
\[ \Rightarrow \angle BOC = \frac{2}{3} \times 180°= 120°\]
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