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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