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Question
If tangents PA and PB from a point P to a circle with centre O are drawn so that ∠APB = 80° then ∠POA = ?

Options
40°
50°
80°
60°
MCQ
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Solution
50°
Explanation:
OP bisects the angle between the equal tangents, so ∠OPA = `(80^circ)/2` = 40°.
In triangle POA, OA ⟂ PA so ∠OAP = 90°.
Thus ∠POA = 90° – 40° = 50°.
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