If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80°, then ∠POA is equal to

1. 50°

2. 60°

3. 70°

4. 80°

We know that,

The lengths of tangents drawn from an external point to a circle are equal.

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

In Δ OAP and Δ OBP,

⇒ OA = OB (radii of the circle are always equal)

⇒ AP = BP (length of the tangents)

⇒ OP = OP (common)

Therefore, by SSS congruency Δ OAP ≅ Δ OBP

Also,

By C.P.C.T.,

⇒ ∠POA = ∠POB

⇒ ∠OPA = ∠OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB.

Hence, ∠OPA = ∠OPB = $\dfrac{1}{2} \times$ ∠APB

= $\dfrac{1}{2}$ × 80°

= 40°.

In Δ OAP

We know that,

OA ⊥ AP ( The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

∴ ∠OAP = 90°.

By angle sum property of a triangle,

⇒ ∠OAP + ∠POA + ∠OPA = 180°

⇒ 90° + ∠POA + 40° = 180°

⇒ 130° + ∠POA = 180°

⇒ ∠POA = 180° - 130°

⇒ ∠POA = 50°.

Hence, Option 1 is the correct option.