Question

In the given figure, AB is tangent to the circle with centre O. If OCB is a straight line segment, the angle BAC is:

Options

• 40°

• 55°

• 35°

• 20°

Answer 1

35°

Explanation:

We know that,

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

∴ OA ⊥ AB

∴ ∠OAB = 90°

Let, ∠BAC = x

From figure,

In △OAC,

∠A = ∠OAB − ∠BAC = 90° − x

Also,

OA = OC (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠C = ∠A = 90° − x

By angle sum property of triangle,

⇒ ∠A + ∠O + ∠C = 180°

⇒ 90° − x + ∠O + 90° − x = 180°

⇒ ∠O + 180° − 2x = 180°

⇒ ∠O = 180° − 180° + 2x = 2x

In △OAB,

By angle sum property of triangle,

⇒ ∠O + ∠A + ∠B = 180°

⇒ ∠O + ∠OAB + ∠B = 180°

⇒ 2x + 90° + 20° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 180° − 110°

⇒ 2x = 70°

⇒ x = `(70°)/2`​ = 35°

⇒ ∠BAC = 35°