BC is a tangent to the circle with center O. OD is radius of the circle. If ∠DOC = 100°, ∠B is equal to :

1. 50°

2. 60°

3. 40°

4. 70°

From figure,

OD = OA (Radius of same circle)

In △OAD,

∠ODA = ∠OAD = x (let) (As angles opposite to equal sides are equal)

Since, exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ ∠DOC = ∠ODA + ∠OAD

⇒ 100° = 2x

⇒ x = $\dfrac{100°}{2}$

⇒ x = 50°.

⇒ ∠OAD = 50°.

We know that,

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

∴ ∠BCA = 90°.

In △ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠BCA + ∠CAB = 180°

⇒ ∠ABC + ∠BCA + ∠OAD = 180° [∵ From figure, ∠CAB = ∠OAD]

⇒ ∠ABC + 90° + 50° = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140° = 40°.

Hence, Option 3 is the correct option.