P is a point in the exterior of odot(O, r) an | vedclass

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(A) $\overline{OX}$ is a radius and $\overleftrightarrow{PX}$ is a tangent at point $X$.According to the theorem,the tangent at any point of a circle

Vedclass · Accepted Answer

$25$

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(A) $\overline{OX}$ is a radius and $\overleftrightarrow{PX}$ is a tangent at point $X$.According to the theorem,the tangent at any point of a circle is perpendicular to the radius through the point of contact.Therefore,$\overline{OX} \perp \overleftrightarrow{PX}$,which implies $m\angle OXP = 90^\circ$.In the right-angled triangle $\Delta PXO$,the sum of the angles is $180^\circ$.Therefore,$m\angle OXP + m\angle XPO + m\angle XOP = 180^\circ$.$90^\circ + 65^\circ + m\angle XOP = 180^\circ$.$155^\circ + m\angle XOP = 180^\circ$.$m\angle XOP = 180^\circ - 155^\circ = 25^\circ$.

$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $m\angle XOP$,if $m\angle XPO = 65^\circ$. (in $^\circ$)

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