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Question
In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60°, calculate:
- ∠QAB,
- ∠PAD,
- ∠CDB.
Solution
i. PAQ is a tangent and AB is the chord.
∠QAB = ∠ADB = 30° ...(Angles in the alternate segment)
ii. OA = OD ...(Radii of the same circle)
∴ ∠OAD = ∠ODA = 30°
But, OA ⊥ PQ
∴ ∠PAD = ∠OAP – ∠OAD
= 90° – 30°
= 60°
iii. BD is the diameter.
∴ ∠BCD = 90° ...(Angle in a semi-circle)
Now in ΔBCD,
∠CDB + ∠CBD + ∠BCD = 180°
`=>` ∠CDB + 60° + 90° = 180°
`=>` ∠CDB = 180° – (60° + 90°)
`=>` ∠CDB = 180° – 150° = 30°
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