Question

In the figure, $PA$ and $PB$ are the two tangents drawn to the circle. $O$ is the center of the circle. $A$ and $B$ are the points of contact of the tangents $PA$ and $PB$ to the circle. If $\angle OPA={35}^o$, then $\angle POB$ is

• A. ${70}^o$
• B. ${55}^o$
• C. ${65}^o$
• D. ${85}^o$

Answer 1

The correct option is B ${55}^o$

Given- $O$ is the centre of a circle to which $PA\&PB$ are two tangents drawn from a point $P$ at $A\&B$ respectively. $\angle APO={35}^o$.
To find out- $\angle POB=?$
Solution- $\angle OAP={90}^o=\angle OBP$ since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Also $PA=PB$ since the lengths of the tangents, drawn from a point to a circle, are equal.
So, between $\Delta POB$ & $\Delta POA$, we have
$PA=PB$,
$PO$ common,
$\angle OAP=\angle OBP$.
$therefore\Delta POB\cong \Delta POA⟹\angle BPO=\angle APO={35}^o$.(by $SAS$ test) .
So, in $\Delta POB$, we have $\angle POB={180}^o-{90}^o-{35}^o={55}^o$.
Ans- Option B.