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In the given figure, `O` is the centre of a circle, `BOA` is its diameter and the tangent at the point `P` meets `BA` extended at `T`. If `/_PBO=30^(@)` then `/_PTA=?`

A

`60^(@)`

B

`30^(@)`

C

`15^(@)`

D

`45^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`/_APB=90^(@)` [angle in a semicircle]
`:./_PAB=90^(@)-/_PBA=90^(@)-30^(@)=60^(@)`
`/_PAT+/_PAB=180^(@)`[ linear pair ]
`implies/_PAT=180^(@)-/_PAB=180^(@)-60^(@)=120^(@)`
`/_APT=/PBA=30^(@)` [angles an alternate segments].
Now, `/_APT+/_PAT+/_PTA=180^(@)` [in `DeltaPAT`]
`implies/_PTA=180^(@)-(30^(@)+120^(@))=30^(@)`.
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