SSC CGL 2024 Tier-I · 23 September 2024, Shift 2 · Quantitative Aptitude · Question 16 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 10 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.

AB is the chord of the circle with centre O. A line segment DOC originating from a point D on the circumference of the circle in major segment meets AB produced at C such that BC = OD. If angle BCO = 30°, then angle AOD is:

Answer

Why

Correct — B. Draw it first: A and B on the circle, C on AB produced beyond B, D on the major arc, with D, O and C in a straight line. Then join OB.

OB and OD are radii and the stem gives BC = OD, so BC = OB.

Triangle OBC is isosceles with BC = OB, so ∠BOC = ∠BCO = 30°.

∠OBA is the exterior angle of triangle OBC at B: ∠OBA = 30° + 30° = 60°.
OA = OB are radii, so triangle OAB is isosceles and ∠OAB = ∠OBA = 60°.

In triangle OAC: ∠AOC = 180° − 60° − 30° = 90°. D, O, C are collinear, so ∠AOD = 180° − 90° = 90° → option (b).

Why the others are wrong

(a)
30° is the given ∠BCO itself. The equal radii build two isosceles triangles that triple it before it reaches the centre.
(c)
60° is ∠OBA, the exterior angle of triangle OBC, and equally ∠OAB — one step short of the angle asked for.
(d)
80° cannot follow from ∠BCO = 30°. Once BC equals the radius the whole figure is fixed, and ∠AOD works out at three times 30°.

Concept

The engine of this configuration is that BC equals a radius, which manufactures a second isosceles triangle where you would not otherwise have one.

OB = BC makes ∠BOC = ∠BCO. The exterior angle of that triangle then doubles the 30°, and OA = OB carries the doubled angle across to ∠OAB.

The angle sum of triangle OAC finishes the job. In general, whenever BC equals the radius in this figure, ∠AOD = 3 × ∠BCO — the classical construction for trebling an angle.

The question prints no diagram, only the four option images, so the marks must be put on your own sketch before anything can be chased.

Key facts

An exterior angle of a triangle equals the sum of the two interior angles opposite it.
Any two radii and the chord joining their ends form an isosceles triangle.
When a segment through the centre meets a chord produced at C with BC equal to the radius, ∠AOD = 3 × ∠BCO.
Here ∠BCO = 30° therefore gives ∠AOD = 90°.

Study next

Angle chasing where every radius is marked equal before anything elseThe angle at the centre is twice the angle at the circumference on the same arcThe alternate segment theorem, for figures with a tangent instead of a secant

Common traps

Stopping at ∠OBA = 60° and picking that.
Forgetting that D, O and C are collinear, which is what converts ∠AOC into ∠AOD.
Reading AB produced as a point between A and B rather than beyond B.

Circle items reward marking the radii before computing anything.

The central-angle to inscribed-angle step is asked at 24 Sep 2024, 12:30, Quant Q.24, where an 80° angle at the centre gives 40° at the circumference, and Quant Q.17 of this paper asks in words for the distance from the centre to the longest chord.

Related PYQs

No directly related past PYQ was found.