Mathematics
In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB.
Circles
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Answer
(i) ∠QAB = ∠ADB [∵ angles in alternate segment are equal.]
∴ ∠QAB = 30°.
Hence, the value of ∠QAB = 30°.
(ii) In △DAO,
OA = OD [∵ radii of the same circle]
So, ∠OAD = ∠ODA = 30° [∵ angles opposite to equal sides are equal.]
We know that,
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OAP = 90°.
From figure,
⇒ ∠PAD = ∠OAP - ∠OAD = 90° - 30° = 60°.
Hence, ∠PAD = 60°.
(iii) In △BCD,
∠BCD = 90° [∵ angle in a semi-circle is a right angle.]
∠CBD = 60°
∠CDB + ∠CBD + ∠BCD = 180° [By angle sum property of triangle]
⇒ ∠CDB + 60° + 90° = 180°
⇒ ∠CDB = 180° - 150° = 30°.
Hence, ∠CDB = 30°.
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