Q3:

Choose the correct option and give justification. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^{\circ}$, then $\angle POA$ is equal to

Given:

1. A circle with center $O$.

2. Two tangents $PA$ and $PB$ drawn from an external point $P$ to the circle.

3. The angle between the tangents, $\angle APB = 80^{\circ}$.

To Find:

The measure of $\angle POA$.

Step 1: Analyzing the Geometry of the Quadrilateral

Consider the quadrilateral $OAPB$. In this figure:

1. $OA \perp PA$ [Since the tangent at any point of a circle is perpendicular to the radius through the point of contact]. Thus, $\angle OAP = 90^{\circ}$.

2. $OB \perp PB$ [Since the tangent at any point of a circle is perpendicular to the radius through the point of contact]. Thus, $\angle OBP = 90^{\circ}$.

Step 2: Calculating the sum of angles in the quadrilateral

The sum of the interior angles of a quadrilateral is $360^{\circ}$.

$\angle OAP + \angle OBP + \angle APB + \angle AOB = 360^{\circ}$

Substituting the known values:

$90^{\circ} + 90^{\circ} + 80^{\circ} + \angle AOB = 360^{\circ}$

$260^{\circ} + \angle AOB = 360^{\circ}$

$\angle AOB = 360^{\circ} - 260^{\circ} = 100^{\circ}$

Step 3: Using Congruency to find $\angle POA$

Consider $\triangle OAP$ and $\triangle OBP$:

1. $OA = OB$ (Radii of the same circle)

2. $OP = OP$ (Common side)

3. $PA = PB$ (Tangents drawn from an external point to a circle are equal in length)

By SSS congruency criterion, $\triangle OAP \cong \triangle OBP$.

By CPCT (Corresponding Parts of Congruent Triangles), $\angle POA = \angle POB$.

Since $\angle AOB = \angle POA + \angle POB$, we have:

$\angle AOB = 2 \times \angle POA$

$100^{\circ} = 2 \times \angle POA$

$\angle POA = \frac{100^{\circ}}{2} = 50^{\circ}$

Final Answer: The value of $\angle POA$ is $50^{\circ}$.