Q12
1 markMCQSection A

  1. If TP and TQ are two tangents to a circle with centre O from an external point T so that ∠POQ=120°\angle POQ = 120°, then ∠PTQ is equal to : (a) 60° (b) 70° (c) 80° (d) 90°

Circles
Angle between two tangents

Options

(A)60°60°
(B)70°70°
(C)80°80°
(D)90°90°
Official Answer

(a) 60°60° — using ∠PTQ+∠POQ=180°\angle PTQ + \angle POQ = 180°, ∠PTQ=180°−120°=60°\angle PTQ = 180° - 120° = 60°.

tangents from external pointtangent perpendicular to radius∠POQ 120°∠PTQ 60°quadrilateral angle sumsupplementary angles

Marking Scheme

  • 11 mark: correct option (a) 60°60°.
  • 2Internal justification: OPTQ angle sum 360° with two 90° angles ⇒ ∠PTQ=360−90−90−120=60°\angle PTQ = 360 - 90 - 90 - 120 = 60°, or the supplementary rule 180−120=60°180 - 120 = 60°.

Hint

Use ∠PTQ+∠POQ=180°\angle PTQ + \angle POQ = 180° (the two tangent–radius right angles make them supplementary), so ∠PTQ=180°−120°\angle PTQ = 180° - 120°.

Quick Oral Answer

Since each radius meets its tangent at 90 degrees, in quadrilateral OPTQ the angle at T plus the 120-degree angle at O equals 180 degrees, so angle PTQ is 60 degrees.

Analysis & Explanation

Since tangents are perpendicular to the radius at the point of contact, quadrilateral OPTQ has two 90° angles, so ∠PTQ=360°−90°−90°−120°=60°\angle PTQ = 360° - 90° - 90° - 120° = 60°.


Concept

  • Tangent–radius property: OP⊥TPOP \perp TP and OQ⊥TQOQ \perp TQ, so ∠OPT=∠OQT=90°\angle OPT = \angle OQT = 90°.
  • Angle sum of quadrilateral OPTQ = 360°.

Key steps

  • ∠PTQ=360°−90°−90°−120°=60°\angle PTQ = 360° - 90° - 90° - 120° = 60° → option (a).
  • Shortcut: ∠PTQ and ∠POQ are supplementary, so ∠PTQ=180°−120°=60°\angle PTQ = 180° - 120° = 60°.

Common mistakes

  • (d) 90°: would only hold if ∠POQ were also 90° — a special case, not this one.
  • (b) 70° and (c) 80°: numerical distractors that don't satisfy the supplementary relation 180° − 120°.
  • Forgetting that BOTH radius–tangent angles are 90°, or mis-adding the quadrilateral's angles.

Real-world

  • The tangent-pair angle rule (∠PTQ + ∠POQ = 180°) is a frequently reused shortcut in circle geometry problems involving external points and tangents.

Common Mistakes

  1. 1Forgetting that both OP⊥TPOP\perp TP and OQ⊥TQOQ\perp TQ, so only counting one 90° angle in the quadrilateral.
  2. 2Wrongly assuming ∠PTQ\angle PTQ equals ∠POQ\angle POQ (120°120°) instead of being supplementary to it.
  3. 3Arithmetic slip when subtracting from 360°, giving 70° or 80° instead of 60°.

Interesting Facts

The tangent–radius perpendicularity is the reason a bicycle chain leaves a sprocket tangentially, and why a stone whirled on a string flies off at a tangent when released.

The two tangents drawn from an external point to a circle are always equal in length (TP=TQTP = TQ), making triangle TPQ isosceles — a fact often paired with this angle result.

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Frequently Asked Questions

Why are ∠PTQ and ∠POQ supplementary?

In quadrilateral OPTQ the two radius–tangent angles are each 90°, so together they use 180° of the 360° total. The remaining 180° is split between ∠POQ and ∠PTQ, making them supplementary.

What property makes ∠OPT and ∠OQT equal to 90°?

A tangent to a circle is always perpendicular to the radius drawn to the point of contact. So OP⊥TPOP\perp TP and OQ⊥TQOQ\perp TQ, giving two right angles.

What if ∠POQ were 90° instead?

Then ∠PTQ=180°−90°=90°\angle PTQ = 180° - 90° = 90° as well, and OPTQ would be a square. That is the case corresponding to distractor (d).