Question

If tangent $PA$ and $PB$ from a point $P$ to a circle with center $O$ are inclined to each other at an angle of $80°$, then $\angle POA$ is equal to

• A. $50°$
• B. $60°$
• C. $70°$
• D. $80°$

Answer 1

The correct option is A

$50°$

Step 1: Using angle sum property of quadrilateral

From the given quadrilateral $OAPB$

$OA$ is perpendicular to $PA$ and $OB$ is perpendicular to $PB$

$\angle A=90°,\angle B=90°$

$\angle P=80°$

The sum of the angles of a quadrilateral is $360°$(Angle sum property of quadrilateral)

Hence

$$\begin{gathered} \angle A+\angle B+\angle P+\angle O=360° \\ \Rightarrow 90°+90°+80°+\angle O=360° \\ \Rightarrow \angle O=100°............\left(1\right) \end{gathered}$$

Step 2: Find the required angle using the congruency of two triangles

Considering $△OAP$ AND $△OBP$

$OA=OB.........$$\left(\mathrm{radius}\mathrm{of}\mathrm{circle}\right)$

$$\begin{gathered} AP=BP \\ \angle OAP=\angle OBP \\ \therefore △OAP=△OBP \\ \therefore △OAP\cong △OBP\left[\mathrm{By}\mathrm{SAS}\mathrm{Congruency}\right] \end{gathered}$$

So, $\angle POA=\angle BOP\left[ByC.P.C.T\right]........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get to know that

$$\begin{gathered} \angle POA+\angle BOP=100° \\ \Rightarrow \angle POA=50° \end{gathered}$$

Hence, $\angle POA$ is $50°$, so, option A is correct