Question

In the given figure, BT and CT′ are two tangents at points B and C on the circle and ∠BPC = 80°. If O is the centre of the circle, then ∠BAC is equal to:

• A. 80°
• B. 60°
• C. 50°
• D. 40°

Answer 1

Answer:

C. 50°

Join BO and OC.

In quadrilateral BOCP,

∠OBP = ∠OCP = 90 (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

∠BPC = 80°

∠BOC = 360 – (90+90+80) = 100°

<BAC=1/2(BOC)=50