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PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 142°, then ∠OAB is equal to:
Answer & Solution
Correct Answer:
Option
C
$$\angle {\text{OAB}} = \frac{{{{180}^ \circ } - {{38}^ \circ }}}{2} = {71^ \circ }$$
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