Mathematics
In the adjoining figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30°, find
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Circles
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Answer
(i) ∠BCO = ∠ACO = 30° (∵ C is the intersecting point of tangent AC and BC. So, OC divides ∠ACB in two halves.)
Hence, the value of ∠BCO = 30°.
(ii) We know that the tangent at any point of a circle and the radius through the point are perpendicular to each other.
∴ ∠OAC = ∠OBC = 90°.
∴ ∠AOC = ∠BOC (∵ tangents are equally inclined to the line joining the point and the centre of the circle.)
Since sum of angles in a triangle = 180.
In AOC
⇒ ∠AOC + ∠OAC + ∠ACO = 180°
⇒ ∠AOC + 90° + 30° = 180°
⇒ ∠AOC + 120° = 180°
⇒ ∠AOC = 180° - 120° = 60°.
∴ ∠BOC = 60°.
From figure,
∠AOB = ∠AOC + ∠BOC = 60° + 60° = 120°.
Hence, the value of ∠AOB = 120°.
(iii) Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.
∴ ∠AOB = 2∠APB (∵ angle subtended at centre by an arc is double the angle subtended at remaining point of the circle.)
∠APB = x ∠AOB = x 120° = 60°.
Hence, the value of ∠APB = 60°.
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