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In the given figure O is center, PQ is tangent at point A. BD is diameter and ∠AOD = 84° then angle QAD is :

  1. 32°

  2. 84°

  3. 48°

  4. 42°

In the given figure O is center, PQ is tangent at point A. BD is diameter and ∠AOD = 84° then angle QAD is : Tangents and Intersecting Chords, Concise Mathematics Solutions ICSE Class 10.

Circles

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Answer

In △OAD,

OA = OD (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠A = ∠D = x (let)

⇒ ∠O + ∠A + ∠D = 180° (By angle sum property of triangle)

⇒ 84° + x + x = 180°

⇒ 2x = 180° - 84°

⇒ 2x = 96°

⇒ x = 96°2\dfrac{96°}{2} = 48°.

From figure,

∠OAD = ∠A = 48°

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAQ = 90°

From figure,

∠DAQ = ∠OAQ - ∠OAD = 90° - 48° = 42°.

Hence, Option 4 is the correct option.

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