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In the given figure, PA and PB are tangents to a circle centred at O. If $\angle AOB = 130^\circ$, then $\angle APB$ is equal to :
- (a)$130^\circ$
- (b)$50^\circ$
- (c)$120^\circ$
- (d)$90^\circ$
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Sol. (B) $50^\circ$ (1 Mark)