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From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ∠APB is (SSC CHSL 2013)
- a)60°
- b)45°
- c)90°
- d)30°
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
From a point P, two tangents PA and PB are drawn to a circle with cent...
AP and PB are two tangents to the circle
∴ ∠OAP = ∠OBP = 90°
In ΔOAP, Let ∠OPA = θ.
OP = 2 × radius {given}
∴ OP = 2 × OA
∴ sinθ = sin 30° ⇒ θ = 30°
Again In ΔBOP, ∠OPA = ∠OPB = θ = 30° {By symmetry}
∴ ∠APB = 30° + 30° = 60°.
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From a point P, two tangents PA and PB are drawn to a circle with cent...
We can conclude that triangle OPA is isosceles because OP is equal to the diameter of the circle. Therefore, angle OAP is equal to angle OPA.
Similarly, triangle OPB is also isosceles because OP is equal to the diameter of the circle. Therefore, angle OBP is equal to angle OPB.
Since the sum of the angles in a triangle is 180 degrees, we can conclude that angle OAP + angle OPA + angle OBP + angle OPB = 180 degrees.
Substituting the equal angles, we have angle OAP + angle OAP + angle OPB + angle OPB = 180 degrees.
Simplifying, we have 2 * angle OAP + 2 * angle OPB = 180 degrees.
Dividing both sides by 2, we have angle OAP + angle OPB = 90 degrees.
Therefore, the sum of the angles between the two tangents from point P to the circle is 90 degrees.
Similarly, triangle OPB is also isosceles because OP is equal to the diameter of the circle. Therefore, angle OBP is equal to angle OPB.
Since the sum of the angles in a triangle is 180 degrees, we can conclude that angle OAP + angle OPA + angle OBP + angle OPB = 180 degrees.
Substituting the equal angles, we have angle OAP + angle OAP + angle OPB + angle OPB = 180 degrees.
Simplifying, we have 2 * angle OAP + 2 * angle OPB = 180 degrees.
Dividing both sides by 2, we have angle OAP + angle OPB = 90 degrees.
Therefore, the sum of the angles between the two tangents from point P to the circle is 90 degrees.
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From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ∠APB is (SSC CHSL 2013)a)60°b)45°c)90°d)30°Correct answer is option 'A'. Can you explain this answer?
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