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Question
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60° ; calculate : ∠CDB
Solution
BD is the diameter.
∴ ∠ BCD = 90° (angle in a semi – circle)
Now in ΔBCD,
∠ CDB + ∠ CBD + ∠ BCD =180°
⇒ ∠ CDB + 60° + 90° =180°
⇒ ∠ CDB =180° - 150° = 30°
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