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

(A) 50° (B) 60° (C) 70° (D) 80°

**Solution:**

Let's draw a figure as per the question.

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 ΔOAP and in ΔOBP

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

AP = BP (length of the tangents)

OP = OP (common)

Therefore, by SSS congruency ΔOAP ≅ ΔOBP

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent then their corresponding parts are equal.

Hence,

∠POA = ∠POB

∠OPA = ∠OPB

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

Hence, ∠OPA = ∠OPB = 1/2 (∠APB )

= 1/2 × 80°

= 40°

By angle sum property of a triangle,

In ΔOAP

∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP (Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 40° = 180°

130° + ∠POA = 180°

∠POA = 180° - 130°

∠POA = 50°

Thus, option (A) 50° is the correct answer.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 10

**Video Solution:**

## If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50° (B) 60° (C) 70° (D) 80°

Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 3

**Summary:**

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to 50°.

**☛ Related Questions:**

- Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
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