Question 3

We will use properties of tangents and quadrilaterals to find the required angle.

Step 1 — Find angles in the quadrilateral

We know that the radius is perpendicular to the tangent.

So, $\angle \text{OAP}$ is a right angle.

This means $\angle \text{OAP} = \mathbf{90^\circ}$.

Also, $\angle \text{OBP}$ is a right angle.

This means $\angle \text{OBP} = \mathbf{90^\circ}$.

We are given that $\angle \text{APB} = \mathbf{80^\circ}$.

Let's look at the quadrilateral AOBP.

Angle Sum of a Quadrilateral: The four interior angles of any quadrilateral always add up to 360°.

The sum of angles in a quadrilateral is $\mathbf{360^\circ}$.

So, we can write the equation:

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

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

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

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

$\boxed{\angle \text{AOB} = 100^\circ}$

Step 2 — Use congruence to find the required angle

Now, let's consider $\triangle \text{OPA}$ and $\triangle \text{OPB}$.

PA and PB are tangents from point P.

Tangents from an External Point are Equal: Any two tangents drawn from the same external point to a circle are always equal in length. This is a standard theorem for tangents.

We know that tangents from an external point are equal in length.

So, $\text{PA} = \text{PB}$.

OA and OB are radii of the same circle.

So, $\text{OA} = \text{OB}$.

OP is a common side to both triangles.

So, $\text{OP} = \text{OP}$.

SSS Congruence Rule: If all three sides of one triangle are equal to the corresponding three sides of another triangle, the triangles are congruent.

By SSS congruence rule, $\triangle \text{OPA} \cong \triangle \text{OPB}$.

CPCT (Corresponding Parts of Congruent Triangles): When two triangles are congruent, all their corresponding angles and sides are equal.

Congruent triangles have equal corresponding parts.

Therefore, $\angle \text{POA} = \angle \text{POB}$.

We found $\angle \text{AOB} = \mathbf{100^\circ}$ in Step 1.

Since $\angle \text{POA} + \angle \text{POB} = \angle \text{AOB}$, and they are equal.

$\angle \text{POA}$ is half of $\angle \text{AOB}$.

$\angle \text{POA} = \frac{1}{2} \times 100^\circ$

$\boxed{\angle \text{POA} = 50^\circ}$

Answer

(iii) The correct option is (A) $50^\circ$.