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

1. 40°

2. 55°

3. 35°

4. 20°

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 = $\dfrac{70°}{2}$ = 35°.

⇒ ∠BAC = 35°.

Hence, Option 3 is the correct option.