AB is a chord in the minor segment of a circle with center O. C is a point between A and B on the minor arc AB. The tangents to the circle at A and B meet at the point D. If ∠ACB = 116°, then the measure of ∠ADB is

Question

Question

This question was previously asked in

SSC CGL 2021 Tier-I (Held On : 20 April 2022 Shift 2)

Answer 1

Given:

F4 Savita SSC 16-11-22 D1 V2

∠ACB = 116°

Concept Used:

When a quadrilateral is inscribed in a circle, the opposite angles of it are supplementary angles.

The angle subtended at the center is always double the angle subtended at the remaining arc.

The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Calculation:

Point P is taken on the major arc of the circle.

Then, A & P, B & P, C & B, and A & C are joined.

From cyclic quadrilateral APBC

∠ACB = 116°

Now, ∠APB = (180 – 116)° = 64°

Now, ∠AOB = (64 × 2)°

⇒ 128°

Since OA = OB = radius of the circle

So, from quadrilateral AOBD

∠ADB = 360° - (∠OBD + ∠OAD + ∠AOB)

⇒ ∠ADB = 360° - (90° + 90° + 128°)

⇒ ∠ADB = 360° - 308°

⇒ ⇒ ∠ADB = 52°

∴ The required measure of ∠ADB is 52°.

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