overline{PA} is a tangent to odot(O, 8) drawn | vedclass

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(D) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$. Therefore,$m\angle OAP = 90^\circ$.Given $m\angl

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$8$

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(D) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$. Therefore,$m\angle OAP = 90^\circ$.Given $m\angle AOP = 45^\circ$.In $\Delta OAP$,the sum of angles is $180^\circ$.$m\angle OPA = 180^\circ - (90^\circ + 45^\circ) = 180^\circ - 135^\circ = 45^\circ$.Since $m\angle AOP = m\angle OPA = 45^\circ$,$\Delta OAP$ is an isosceles triangle.Therefore,the sides opposite to equal angles are equal: $AP = OA$.Since $OA$ is the radius of the circle,$OA = 8$.Thus,$AP = 8$.

$\overline{PA}$ is a tangent to $\odot(O, 8)$ drawn from a point $P$ outside the circle. If $m\angle AOP = 45^\circ$,then $AP = \ldots$

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