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Question
PA and PB are tangents to a circle with centre O. If angle BPA = 70°, the angle ACB is:
Options
70°
105°
140°
55°
MCQ
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Solution
55°
Explanation:
Join OA and OB.
We know that,
Tangent at any point of a circle and the radius through this point are perpendicular to each other.
∴ ∠OAP = 90° and ∠OBP = 90°
In quadrilateral OAPB,
⇒ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
⇒ 90° + 70° + 90° + ∠BOA = 360°
⇒ ∠BOA + 250° = 360°
⇒ ∠BOA = 360° − 250° = 110°
We know that,
The angle which an arc of a circle subtends at the center is double that which it subtends at any point on the remaining part of the circumference.
∴ ∠AOB = 2∠ACB
⇒ ∠ACB = `1/2 ∠AOB`
⇒ ∠ACB = `1/2×110°` = 55°
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