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(D) $1$. In $	riangle OBP$,since $PB$ is a tangent to the circle at $B$,the radius $OB$ is perpendicular to the tangent $PB$. Thus,$\angle OBP = 90^\

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(D) $1$. In $	riangle OBP$,since $PB$ is a tangent to the circle at $B$,the radius $OB$ is perpendicular to the tangent $PB$. Thus,$\angle OBP = 90^\circ$.$2$. In the right-angled triangle $	riangle OBP$,the sum of angles is $180^\circ$. Therefore,$\angle POB + \angle OBP + \angle OPB = 180^\circ$.$3$. Substituting the known values: $\angle POB + 90^\circ + 35^\circ = 180^\circ$,which gives $\angle POB = 180^\circ - 125^\circ = 55^\circ$.$4$. Since the tangents from an external point are equally inclined to the line joining the center to that point,$	riangle OAP \cong 	riangle OBP$. Thus,$\angle AOP = \angle POB = 55^\circ$.$5$. Therefore,$m\angle AOB = \angle AOP + \angle POB = 55^\circ + 55^\circ = 110^\circ$.

$\stackrel{\leftrightarrow}{PA}$ and $\stackrel{\leftrightarrow}{PB}$ are tangents to the circle $\odot(O, r)$ at points $A$ and $B$ respectively. If $m\angle OPB = 35^\circ$,then $m\angle AOB = \ldots$ (in $^\circ$)

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