Question 4 of 11intermediate🔧 ApplyMCQ1 mark

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

Exercise: EXERCISE 10.2 | Q: 3 | (Page 8)
For More Understanding

Explanation

Option A is correct. Using Theorem 10.1 (tangent perpendicular to radius) and Theorem 10.2 (tangents from external point are equal), we can show that triangles OAP and OBP are congruent, leading to OP bisecting the angle between the tangents. Hence, ∠OPA = 40°. Then in right triangle OAP, ∠POA = 90° - 40° = 50°.

Solution Steps

  1. Step 1: PA = PB (Theorem 10.2) → ΔPAB is isosceles.

  2. Step 2: OA ⟂ PA and OB ⟂ PB (Theorem 10.1) → ∠OAP = ∠OBP = 90°.

  3. Step 3: In ΔOAP and ΔOBP, OA = OB (radii), OP common, ∠OAP = ∠OBP = 90° → ΔOAP ≅ ΔOBP (RHS). Hence, ∠OPA = ∠OPB and ∠POA = ∠POB.

  4. Step 4: Since ∠OPA = ∠OPB, OP bisects ∠APB. Given ∠APB = 80°, so ∠OPA = 40°.

  5. Step 5: In right ΔOAP, ∠POA + ∠OPA = 90° → ∠POA = 90° - 40° = 50°.

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