Write 'True' or 'False' and give reasons for your answer.
In the figure,$BOA$ is a diameter of a circle and the tangent at a point $P$ meets $BA$ extended at $T$. If $\angle PBO = 30^{\circ}$,then $\angle PTA$ is equal to $30^{\circ}$.
In the figure,$BOA$ is a diameter of a circle and the tangent at a point $P$ meets $BA$ extended at $T$. If $\angle PBO = 30^{\circ}$,then $\angle PTA$ is equal to $30^{\circ}$.
(A) True.
In $\triangle PBO$,$OB = OP$ (radii of the same circle).
Therefore,$\angle OPB = \angle PBO = 30^{\circ}$.
By angle sum property in $\triangle PBO$,$\angle POB = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.
Since $BOA$ is a straight line,$\angle POA = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\triangle OPT$,$OP \perp PT$ (radius is perpendicular to the tangent at the point of contact).
Therefore,$\angle OPT = 90^{\circ}$.
In $\triangle OPT$,$\angle PTA = 180^{\circ} - (\angle POT + \angle OPT) = 180^{\circ} - (60^{\circ} + 90^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Thus,the statement is True.
In $\triangle PBO$,$OB = OP$ (radii of the same circle).
Therefore,$\angle OPB = \angle PBO = 30^{\circ}$.
By angle sum property in $\triangle PBO$,$\angle POB = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.
Since $BOA$ is a straight line,$\angle POA = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\triangle OPT$,$OP \perp PT$ (radius is perpendicular to the tangent at the point of contact).
Therefore,$\angle OPT = 90^{\circ}$.
In $\triangle OPT$,$\angle PTA = 180^{\circ} - (\angle POT + \angle OPT) = 180^{\circ} - (60^{\circ} + 90^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Thus,the statement is True.
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