Question

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of 80^o then $\angle POA$ is equal to

• A. 70^o
• B. 60^o
• C. 80^o
• D. 50^o

Answer 1

$The correct answer is Option D 50^o

Given: $PA$ and $PB$ are two tangents a circle and $\angle APB={80}^o$

Since OA⊥PA and OB⊥PB, Then $\angle OAP={90}^{\circ }$ and $\angle OBP={90}^{\circ }$

In, $\Delta OAP$ and $\Delta OBP$

$OA=OB$(radius)

$OP=OP$(Common)

$PA=PB$ (lengths of tangents drawn from external pointis )

$\therefore \Delta OAP\cong \Delta OBP$ $(SSS$ congruence)

So,[$\angle OPA=\angle OPB$ (byCPCT)]

So, $\angle OPA=\frac{1}{2}\angle APB$

$=\frac{1}{2}$ $\times {80}^{\circ }={40}^{\circ }$

In $\Delta OPA$,

$\angle POA+\angle OPA+\angle OAP={180}^{\circ }$

$\angle POA+{40}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\angle POA+{130}^{\circ }={180}^{\circ }$

$\angle POA={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Hence, the value of $\angle POA$ is ${50}^{\circ }$