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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.

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, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

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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