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(A) In the circle $\odot(O, r)$,$PA$ and $PB$ are tangents drawn from an external point $P$ to the points of contact $A$ and $B$.By the properties of

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) In the circle $\odot(O, r)$,$PA$ and $PB$ are tangents drawn from an external point $P$ to the points of contact $A$ and $B$.By the properties of tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.In the quadrilateral $OAPB$,the sum of the interior angles is $360^\circ$.Thus,$\angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ$.Substituting the given values: $100^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ$.$280^\circ + \angle APB = 360^\circ$,which gives $\angle APB = 80^\circ$.The line $OP$ bisects $\angle APB$. Therefore,$m\angle OPB = \frac{1}{2} \angle APB = \frac{1}{2} 	imes 80^\circ = 40^\circ$.

$\overleftrightarrow{PA}$ and $\overleftrightarrow{PB}$ are tangents to the circle $\odot(O, r)$ at $A$ and $B$ respectively. If $m\angle AOB = 100^\circ$,then $m\angle OPB = \dots$ (in $^\circ$)

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