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In the given figure, `O` is the centre of the circle. `AB` is the tangent to the circle at the point `P`. If `/_PAO=30^(@)` then `/_CPB+/_ACP` is equal to

A

`60^(@)`

B

`90^(@)`

C

`120^(@)`

D

`150^(@)`

Text Solution

Verified by Experts

`/_DPC=90^(@)` [angle in semicircle]
and `/_DPA=/_DCP` [angles in alternate segments]
Now, `/_CPB+/_DPA+/_DPC=180^(@)`
`implies/_CPB+/_DPA=90^(@)implies/_CPB+/_DCP=90^(@)`
`implies/_CPB+/_ACP=90^(@)`
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RS AGGARWAL-CIRCLES-Multiple-choice questions (MCQ)
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