In the adjoining figure, ABCD is a cyclic quadrilateral in which ∠BCD = 100° and ∠ABD = 50°. Find ∠ADB.

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Solution

Given: ABCD is a cyclic quadrilateral.
∴ ∠DAB + ∠DCB = 180° ( Opposite angles of a cyclic quadrilateral are supplementary)
⇒ ∠DAB + 100° = 180°
⇒ ∠DAB = (180° – 100°) = 80°
Now, in ΔABD, we have:
⇒ ∠DAB + ∠ABD + ∠ADB = 180°
⇒ 80° + 50° + ∠ADB = 180°
⇒ ∠ADB = (180° – 130°) = 50°
Hence, ∠ADB = 50°

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