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Two tangents are drawn from a point P to a circle at A and B. O is the centre of the circle. If ∠AOP = 60°, then ∠APB is (SSC CGL 1st Sit. 2012)
- a)120°
- b)90°
- c)60°
- d)30°
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Two tangents are drawn from a point P to a circle at A and B. O is the...
The correct answer to this question is option 'C' (60°).
Explanation:
- To solve this problem, we need to use the property that the angle between a tangent and the radius drawn to the point of contact is 90°.
- Let's denote the angle ∠AOP = 60°, then we have ∠OAP = 90° (as OA is a radius).
- Similarly, ∠OBP = 90° (as OB is a radius).
- Therefore, in triangle OAP, we have ∠OAP + ∠APO + ∠OPA = 180° (sum of angles in a triangle is 180°).
- Substituting the known values, we get 60° + 90° + ∠OPA = 180°.
- Solving for ∠OPA, we find ∠OPA = 30°.
- Similarly, in triangle OBP, we have ∠OBP + ∠BPO + ∠OPB = 180°.
- Substituting the known values, we get 90° + ∠BPO + 60° = 180°.
- Solving for ∠BPO, we find ∠BPO = 30°.
- Finally, in triangle APB, ∠APB = ∠OPA + ∠BPO = 30° + 30° = 60°.
Therefore, the angle ∠APB is 60°.
Explanation:
- To solve this problem, we need to use the property that the angle between a tangent and the radius drawn to the point of contact is 90°.
- Let's denote the angle ∠AOP = 60°, then we have ∠OAP = 90° (as OA is a radius).
- Similarly, ∠OBP = 90° (as OB is a radius).
- Therefore, in triangle OAP, we have ∠OAP + ∠APO + ∠OPA = 180° (sum of angles in a triangle is 180°).
- Substituting the known values, we get 60° + 90° + ∠OPA = 180°.
- Solving for ∠OPA, we find ∠OPA = 30°.
- Similarly, in triangle OBP, we have ∠OBP + ∠BPO + ∠OPB = 180°.
- Substituting the known values, we get 90° + ∠BPO + 60° = 180°.
- Solving for ∠BPO, we find ∠BPO = 30°.
- Finally, in triangle APB, ∠APB = ∠OPA + ∠BPO = 30° + 30° = 60°.
Therefore, the angle ∠APB is 60°.
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Community Answer
Two tangents are drawn from a point P to a circle at A and B. O is the...
In right Δs OAP and OPB,
AP = PB, OA = OB
OP = OP
∴ ΔOAP ≌ ΔOBP
∴ ∠AOP = ∠POB and ∠APO = ∠OPB
From ΔAOP,
∠APO = 180° – 90° – 60° = 30°
∴ ∠APB = 2 × 30 = 60°
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Two tangents are drawn from a point P to a circle at A and B. O is the centre of the circle. If ∠AOP = 60°, then ∠APB is (SSC CGL 1st Sit. 2012)a)120°b)90°c)60°d)30°Correct answer is option 'C'. Can you explain this answer?
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