Question

O is centre of the circle, PB and PC are tangents and ∠BPC = 50°.

Statement (1): ∠BAC = ∠P = 50°

Statement (2): ∠BOC + 50° = 180°

⇒ ∠BOC = 130°

∴ ∠BAC = 65°

विकल्प

• Both the statements are true.

• Both the statements are false.

• Statement 1 is true, and statement 2 is false.

• Statement 1 is false, and statement 2 is true.

Answer 1

Statement 1 is false, and statement 2 is true.

Explanation:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OB ⊥ BP and OC ⊥ CP

⇒ ∠OBP = 90° and ∠OCP = 90°

OCPB is a quadrilateral.

∴ ∠OBP + ∠BPC + ∠OCP + ∠BOC = 360°

⇒ 90° + 50° + 90° + ∠BOC = 360°

⇒ 230° + ∠BOC = 360°

⇒ ∠BOC = 360° − 230°

⇒ ∠BOC = 130°

We know that,

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOC = 2 × ∠BAC

⇒ 130° = 2 × ∠BAC

⇒ ∠BAC = `(130°)/2`​ = 65°

So, Statement 1 is false, and statement 2 is true.