Q13

1 markMCQSection A

In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 125°$ , then ∠APO is equal to : (a) 25° (b) 65° (c) 90° (d) 35°

Q: 13. In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 125°$, then ∠APO is equal to : (a) 25° (b) 65° (c) 90° (d) 35°

(d) $35°$, since $OA \perp PA$ makes $\angle OAP = 90°$, and the angle sum of triangle OAP gives $\angle APO = 180° - 90° - 55° = 35°$.

Circles

Tangent perpendicular to radius

Options

(A)

25°

(B)

65°

(C)

90°

(D)

35°

Official Answer

(d) $35°$ , since $OA \perp PA$ makes $\angle OAP = 90°$ , and the angle sum of triangle OAP gives $\angle APO = 180° - 90° - 55° = 35°$ .

tangentradiusperpendicularpoint of contact∠OAP = 90°angle sum property35 degreesexternal point

Marking Scheme

11 mark for the correct option (d) 35°.
2Working expected: $\angle OAP = 90°$ (radius ⊥ tangent); $\angle AOP = 180° - 125° = 55°$ ; $\angle APO = 180° - 90° - 55° = 35°$ .
3No marks for merely writing 35° with a contradictory reason; accept the answer if radius–tangent perpendicularity is used correctly.

Hint

The radius meets the tangent at 90° ( $\angle OAP = 90°$ ); find ∠AOP using the straight line at O, then use angle sum of triangle OAP.

Quick Oral Answer

Since the radius OA is perpendicular to the tangent PA, ∠OAP is 90°; ∠AOP is 180° minus 125°, which is 55°, so ∠APO is 180° minus 90° minus 55°, giving 35°.

Analysis & Explanation

Tests the tangent–radius perpendicularity theorem to find an unknown angle in a triangle formed by the centre, the point of contact, and the external point.

Concept

The tangent to a circle at any point is perpendicular to the radius drawn to that point of contact, so $OA \perp PA$ and $\angle OAP = 90°$ .
A, O, B lie such that ∠AOP is the supplement of the given ∠POB (linear pair on line AOB), giving $\angle AOP = 180° - 125° = 55°$ .
In triangle OAP, angle sum gives $\angle APO = 180° - 90° - 55° = 35°$ .

Common mistakes

Forgetting the radius–tangent right angle and instead assuming ∠OAP is some other value.
Adding 125° and 90° directly instead of first finding the supplementary angle ∠AOP.

Common Mistakes

1Selecting (c) $90°$ by reporting the radius–tangent angle ∠OAP instead of the required ∠APO.
2Forgetting that ∠POB and ∠AOP are supplementary and using $125°$ directly inside the triangle, giving a negative or impossible angle.
3Assuming triangle OAP is isosceles ( $OA = PA$ ) — OA is a radius and PA is a tangent length, so they are generally unequal.

Interesting Facts

The theorem that a tangent is perpendicular to the radius at the point of contact appears as Theorem 10.1 in the NCERT Class 10 Maths textbook (Chapter 10, Circles).

Because OA ⊥ PA, the point A always lies on the circle with OP as diameter — a neat consequence used to construct tangents from an external point.

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Frequently Asked Questions

Why is $\angle OAP$ equal to $90°$ ?

Because A is the point of contact of the tangent PA, and the tangent at any point of a circle is perpendicular to the radius (OA) drawn to that point. This is a standard circle theorem.

How do we get $\angle AOP = 55°$ from $\angle POB = 125°$ ?

Points A, O and B are arranged so that ∠AOP and ∠POB lie on a straight line at O and are supplementary. Hence $\angle AOP = 180° - 125° = 55°$ .

Could the triangle be solved without the perpendicular property?

No. The right angle at A is what makes triangle OAP solvable with just one more angle; without radius ⊥ tangent you cannot fix any angle of the triangle.