In the figure, O is the centre and BOA is a diameter of the circle. A tangent drawn to the circels at the point P intersects the extended BA at the point T. If `angle PBO=30^(@),` find the value of `angle PAT.`

Explanation

Let us join O, P and A, P. ltbr PT is a tangent to the circel at P and OP is a radius through the point of contact P.
`therefoer OP bot PT`
`therefore angle OPT=90^(@), angle OPB=angle OBP=30^(@)`
`therefore angle BPT=angle OPB +angle OPT`
`=30^(@)+90^(@) [because angle OPB =30^(@)and angle OPT = 90^(@)] =120^(@)`
Now, in `DeltaBPT, angleBTP+angleTPB +angle PBT=180^(@)`
or `angle BPT +120^(@)+30^(@) =180^(@)`
or `anglePAT+150^(@)=180^(@)or, angle PAT=180^(@)-150^(@), or, angle PAT=30^(@)`
`therefore` the value of `angle PTA=30^(@),`