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Question
In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR such that ∠BQR = 70°. Then, AQB = ?
Options
20°
35°
40°
45°
MCQ
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Solution
40°
Explanation:
∠QAB = ∠BQR = 70° ...[Angles in alternate segments]
∠ABQ = ∠BQR = 70° ...[Alternate int. angles]
In ΔAQB, we have
∠AQB = 180° – (∠QAB + ∠ABQ)
= 180° – (70° + 70°)
= 40°
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