| 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\[\angle PBO=30{}^\circ ,\]then \[\angle PTA\]is equal to |
 |
A) \[60{}^\circ \]
B) \[30{}^\circ \]
C) \[15{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: B
Solution :
| Join \[OP,\]\[\angle BPA=90{}^\circ \] [angle in semi-circle] |
| In \[\Delta PBA,\] |
| \[\angle BPA+\angle PBA+\angle BAP=180{}^\circ \] |
| \[\Rightarrow \] \[90{}^\circ +30{}^\circ +\angle BAP=180{}^\circ \] |
| \[\therefore \] \[\angle BAP=60{}^\circ \] |
| But \[BAT\]is a straight angle, |
| \[\Rightarrow \]\[\angle BPA+\angle PAT=180{}^\circ \] |
| \[\Rightarrow \]\[60{}^\circ +\angle PAT=180{}^\circ \] |
| \[\therefore \] \[\angle PAT=120{}^\circ \] |
| \[OA=OP\] [radii] |
| \[\Rightarrow \]\[\angle OPA=\angle OAP=\angle BAP=60{}^\circ \] |
| Now, \[\angle OPT=90{}^\circ \] |
| \[\Rightarrow \]\[\angle OPA+\angle APT=90{}^\circ \] |
| \[60{}^\circ +\angle APT=90{}^\circ \] |
| \[\therefore \] \[\angle APT=30{}^\circ \] |
| In\[\Delta PAT,\]we have |
| \[\angle PAT+\angle APT+\angle PTA=180{}^\circ \] |
| \[\therefore \] \[\angle PTA=30{}^\circ \] |
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