SSC CGL 2024 Tier-I · 9 September 2024, Shift 2 · Quantitative Aptitude · Question 8 of 100

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 5 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.

Let C be a circle with centre O and P be an external point to C. Let PA and PB be two tangents to C with A and B being the points of tangency, respectively. If PA and PB are inclined to each other at an angle of 60°, then find ∠ POA.

(a)80°
(b)30°
(c)40°
(d)60°

Answer

Why

Correct — D. Two standard facts settle the figure.

Rule: a radius meets a tangent at right angles at the point of contact, so ∠OAP = 90°.
Rule: PA = PB (tangents from an external point are equal), so OP is the axis of symmetry and bisects ∠APB.

∠APO = 60° ⁄ 2 = 30°
∠POA = 180° − 90° − 30° = 60° → option (d)

Same answer from the quadrilateral OAPB: its angles at A and B are 90° each, so ∠AOB = 180° − 60° = 120°, and ∠POA is half of that.

Why the others are wrong

(a)
80° — 80° would force ∠APO = 10° and so ∠APB = 20°, contradicting the 60° given. Since ∠OAP = 90°, the other two angles of triangle OAP must add to 90°.
(b)
30° — 30° is ∠APO, the half-angle at P, not the angle at the centre — the two are complementary in right triangle OAP, so the angle at O is 60°.
(c)
40° — ∠POA = 40° would make ∠APO = 50° and therefore ∠APB = 100°, but the question fixes ∠APB at 60°.

Concept

Two tangents from an external point create a kite OAPB: OA = OB as radii, PA = PB as equal tangents, and OP as the axis of symmetry, so OP bisects both ∠APB and ∠AOB.

Everything rests on one fact — a tangent is perpendicular to the radius at the point of contact — which makes OAP and OBP right triangles with the right angle at A and at B, never at P or at O.

Because ∠OAP = 90°, the angles at O and at P are complementary, so knowing either one gives the other immediately and the whole figure follows from the single angle at P.

Here PA = PB with ∠APB = 60° makes triangle APB equilateral, so the chord AB equals the tangent length.

SSC often puts that side result in the next question built on the same figure.

Key facts

A tangent is perpendicular to the radius at the point of contact, so ∠OAP = ∠OBP = 90°.
Tangents from an external point are equal, PA = PB, and OP bisects both ∠APB and ∠AOB.
In quadrilateral OAPB, ∠AOB + ∠APB = 180°.
Here ∠APB = 60°, so ∠AOB = 120°.

Study next

Tangent length from an external point, PA = √(OP² − r²), combined with this angle chaseThe alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segmentCommon tangents to two circles, direct and transverse, and the length of each

Common traps

Forgetting that OP bisects ∠APB and carrying the full 60° into triangle OAP.
Placing the right angle at O or at P instead of at the point of contact A.
Answering ∠AOB, which is 120°, when the question asks for ∠POA.

SSC gives the angle between two tangents and asks for the angle at the centre, for the half-angle at either vertex, or for the tangent length. The single relation ∠AOB = 180° − ∠APB settles most of these in one line.

Related PYQs

No directly related past PYQ was found.