In Q.1 to 3, choose the correct option and give justification. 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of , then is equal to (A) (B) (C) (D)

Understand the Question The tangent to a circle is perpendicular to the radius at the point of contact, meaning . In quadrilateral , the sum of interior angles is , which gives the central angle . The line segment bisects (by congruence of and ), so . Step 1 · Find in Quadrilateral The radius is perpendicular to the tangent at the point of contact. Given . In quadrilateral , the sum of interior angles is : Step 2 · Find Using Congruence In and : (tangents from an external point) (radii of the same circle) (common side) By SSS congruence criterion: By CPCT: Since : Answer > (A) Common Mistakes Confusing with : Forgetting to divide by after calculating the central angle of . Confusing with : , whereas .

Circles | Exercise 10.2

Question 3

In Q.1 to 3, 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 $\angle \text{POA}$ is equal to (A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$

Solution

Understand the Question

The tangent to a circle is perpendicular to the radius at the point of contact, meaning $\angle \text{OAP} = \angle \text{OBP} = 90^\circ$ .
In quadrilateral $\text{OAPB}$ , the sum of interior angles is $360^\circ$ , which gives the central angle $\angle \text{AOB} = 180^\circ - 80^\circ = 100^\circ$ .
The line segment $OP$ bisects $\angle \text{AOB}$ (by congruence of $\triangle \text{OPA}$ and $\triangle \text{OPB}$ ), so $\angle \text{POA} = \dfrac{1}{2}\angle \text{AOB}$ .

Step 1 · Find $\angle \text{AOB}$ in Quadrilateral $\text{OAPB}$

The radius is perpendicular to the tangent at the point of contact. Diagram 1

$\angle \text{OAP} = 90^\circ, \quad \angle \text{OBP} = 90^\circ$

Given $\angle \text{APB} = 80^\circ$ .

In quadrilateral $\text{AOBP}$ , the sum of interior angles is $360^\circ$ :

\begin{aligned} \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 \\ \angle \text{AOB} &= 100^\circ \end{aligned}

Step 2 · Find $\angle \text{POA}$ Using Congruence

In $\triangle \text{OPA}$ and $\triangle \text{OPB}$ :

$\text{PA} = \text{PB}$ (tangents from an external point)
$\text{OA} = \text{OB}$ (radii of the same circle)
$\text{OP} = \text{OP}$ (common side)

By SSS congruence criterion: $\triangle \text{OPA} \cong \triangle \text{OPB}$

By CPCT: $\angle \text{POA} = \angle \text{POB}$

Since $\angle \text{POA} + \angle \text{POB} = \angle \text{AOB} = 100^\circ$ : $\angle \text{POA} = \dfrac{1}{2} \times 100^\circ = 50^\circ$

Answer

(A) $50^\circ$

Common Mistakes

Confusing $\angle \text{POA}$ with $\angle \text{AOB}$ : Forgetting to divide by $2$ after calculating the central angle of $100^\circ$ .
Confusing $\angle \text{POA}$ with $\angle \text{OPA}$ : $\angle \text{OPA} = \dfrac{1}{2} \times 80^\circ = 40^\circ$ , whereas $\angle \text{POA} = 90^\circ - 40^\circ = 50^\circ$ .

Question 3

More questions in Exercise 10.2