SSC CGL 2024 Tier-I · 17 September 2024, Shift 1 · Quantitative Aptitude · Question 16 of 98

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

AP and AQ are two tangents drawn to a circle with center O from an external point A. If ∠ PAQ = 40°, then ∠ POQ is:

(a)120°
(b)140°
(c)130°
(d)150°

Answer

Why

Correct — B. A tangent meets the radius at 90° at the point of contact, so ∠OPA = ∠OQA = 90°.

OPAQ is a quadrilateral, and its four angles sum to 360°:
∠POQ + 90° + 90° + ∠PAQ = 360°
∠POQ = 360° − 180° − 40° = 140°

So the centre angle and the angle between the tangents are supplementary, and 180° − 40° = 140° → option (b).

Why the others are wrong

(a)
120° — 120° pairs with an external angle of 60°, since the two must total 180°. The stem states 40°, so 120° answers a different figure.
(c)
130° — 130° pairs with 50°. Add and check: 130 + 40 = 170, ten degrees short of the 180° the centre and external angles always make.
(d)
150° — 150° pairs with 30°, and 150 + 40 = 190, ten degrees over. The two right angles at P and Q take exactly 180° of the quadrilateral, leaving no room for it.

Concept

Join OP and OQ, the radii to the two points of contact. A radius is perpendicular to the tangent at its point of contact, so both ∠OPA and ∠OQA are right angles.

That turns OPAQ into a quadrilateral with two known right angles. Its angles sum to 360°, and the two right angles use 180° of that.

What is left is the standing result: ∠POQ + ∠PAQ = 180°. Once you carry that, the question is a single subtraction and needs no construction at all.

AP = AQ as well, because tangents drawn from one external point are equal, which makes triangle APQ isosceles. That fact is not needed here but is the usual second half of these items.

Key facts

A tangent is perpendicular to the radius drawn to its point of contact.
For two tangents from an external point A to a circle centred O, ∠POQ + ∠PAQ = 180°.
The two tangent segments from an external point are equal in length: AP = AQ.

Study next

Length of a tangent from an external point, √(d² − r²)The angle in the alternate segmentDirect and transverse common tangents to two circles

Common traps

Assuming ∠POQ equals ∠PAQ, or that it is twice it.
Halving ∠PAQ to 20° and answering about triangle OPA instead of the whole quadrilateral.
Applying the 360° sum to a triangle, which drops one of the two right angles.

SSC gives the angle between the tangents and asks for the angle at the centre, reverses the pair, or asks for the half-angle ∠POA instead.

The same supplementary pair is asked 9 Sep 2024, 12:30, Quant Q.8, where tangents inclined at 60° lead to ∠POA. The right angle at the point of contact drives 26 Sep 2024, 09:00, Quant Q.18, where the area of the tangent quadrilateral is asked.

Related PYQs

No directly related past PYQ was found.