O is the center of the circle and two tangents are drawn from a point P to this circle at points A and B. If ∠AOP = 50°, then what is the value (in degrees) of ∠APB?

Given : $$\angle$$ AOP = $$50^\circ$$

To find : $$\angle$$ APB = $$2\theta$$ = ?

Solution : $$\angle$$ APO = $$\frac{1}{2} \times$$ $$\angle$$ APB

=> $$\angle$$ APO = $$\frac{1}{2} \times 2\theta=\theta$$

Also, the radius of a circle intersects the tangent at the circumference of circle at $$90^\circ$$

=> $$\angle$$ OAP = $$90^\circ$$

In $$\triangle$$ AOP

=> $$\angle$$ AOP + $$\angle$$ APO + $$\angle$$ OAP = $$180^\circ$$

=> $$\theta + 50^\circ+90^\circ=180^\circ$$

=> $$\theta=180^\circ-140^\circ$$

=> $$\theta=40^\circ$$

$$\therefore$$ $$\angle$$ APB = $$2\theta=2\times40=80^\circ$$

=> Ans - (B)