Question

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

विकल्प

• 32°

• 84°

• 48°

• 42°

Answer 1

42°

Explanation:

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`​ = 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°