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Question
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO = 30°, find:
- ∠BCO
- ∠AOB
- ∠APB
Solution
In the given fig, O is the centre of the circle and CA and CB are the tangents to the circle from C. Also, ∠ACO = 30°
P is any point on the circle. PA and PB are joined.
To find:
- ∠BCO
- ∠AOB
- ∠APB
Proof:
i. In ΔOAC and OBC
OC = OC ...(Common)
OA = OB ...(Radius of the circle)
CA = CB ...(Tangents to the circle)
∴ ΔOAC ≅ ΔOBC ...(SSS congruence criterion)
∴ ∠ACO = ∠BCO = 30°
ii. ∠ACB = 30° + 30° = 60°
∴ ∠AOB + ∠ACB = 180°
`=>` ∠AOB + 60° = 180°
`=>` ∠AOB = 180° – 60°
`=>` ∠AOB = 120°
iii. Arc AB subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle.
∴ `∠APB = 1/2 ∠AOB`
= `1/2 xx 120^circ`
= 60°
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