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(C) Given that $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$ from an external point $P$.According to the theorem,

Vedclass · Accepted Answer

$60$

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(C) Given that $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$ from an external point $P$.According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.Therefore,$\overline{ PA } = \overline{ PB }$.In $\Delta PAB$,since $\overline{ PA } = \overline{ PB }$,the triangle is an isosceles triangle.In an isosceles triangle,the angles opposite to the equal sides are equal.Thus,$m \angle PAB = m \angle PBA$.Given that $m \angle PAB = 60^{\circ}$,it follows that $m \angle PBA = 60^{\circ}$.

In the figure,$\overrightarrow{ PA }$ and $\overrightarrow{ PB }$ are tangents to $\odot( O , r)$. If $m \angle PAB = 60^{\circ}$,then $m \angle PBA = \ldots$ (in $^{\circ}$)

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