SSC CGL 2025 Tier-I · 21 September 2025, Shift 3 · Quantitative Aptitude · Question 21 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

If sin(A) = cos(B), cos(A) = sin(C), and A + B + C = 90°, then what is the value of angle A?

(a)90°
(b)60°
(c)30°
(d)15°

Answer

Why

Correct — A. Use complementary angles: sin A = cos(90° − A) and cos A = sin(90° − A).

sin A = cos B, so B = 90° − A
cos A = sin C, so C = 90° − A

Substitute into A + B + C = 90°: A + (90° − A) + (90° − A) = 90°
Simplify: 180° − A = 90°
Solve: A = 90° → option (a)

Why the others are wrong

(b)
60° — 60° makes B = C = 90° − 60° = 30°. The total is then 60° + 30° + 30° = 120°, not the 90° the stem fixes.
(c)
30° — 30° is 90° split three ways, but A = B = 30° breaks sin A = cos B (1⁄2 against √3⁄2). The conditions give B = C = 60° and a total of 150°.
(d)
15° — 15° makes B = C = 75°, so the total is 15° + 75° + 75° = 165°. The sum 180° − A reaches 90° only at A = 90°.

Concept

Complementary angles add to 90°, and the sine of one equals the cosine of the other: sin θ = cos(90° − θ).

So each equation hands you one angle in terms of A: B = 90° − A and C = 90° − A. The third condition, A + B + C = 90°, then becomes 180° − A = 90°.

Check the key: with A = 90° both B and C are 0°. sin 90° = cos 0° = 1, cos 90° = sin 0° = 0, and 90° + 0° + 0° = 90°. The stem never says the angles form a triangle, so 0° is allowed.

The working reads all three as angles from 0° to 90°. Allow a negative angle and A = 30°, B = −60°, C = 120° also fits every condition, so the key rests on that range.

Key facts

sin θ = cos(90° − θ) and cos θ = sin(90° − θ).
For angles from 0° to 90°, sin A = cos B means A + B = 90°.
sin 90° = cos 0° = 1 and cos 90° = sin 0° = 0.

Study next

Equations like sin 3A = cos(A − 26°) solved with complementstan A = cot B, which also means A + B = 90°Rewriting sin 74° as cos 16° to simplify an expression

Common traps

Splitting 90° three ways and answering 30° without using the sine and cosine conditions.
Rejecting 90° because it makes B and C zero, though the stem never says the angles form a triangle.

26 Sep 2024, 09:00, Quant Q.10 gives sin(48° + k) = cos 13° with 48° + k acute. The angles are complementary, so k = 90° − 13° − 48°, keyed 29.

24 Sep 2024, 09:00, Quant Q.22 gives sin 3A = cos(A − 26°) with 3A acute. Then 4A − 26° = 90°, keyed 29°.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 98 SSC CGL questions on it, every solved shift.