SSC CGL 2025 Tier-I · 22 September 2025, Shift 1 · Quantitative Aptitude · Question 24 of 98

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 13 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 sinA = m⁄n and A ∈ (0, π⁄2), then what is cosA in terms of m and n?

(a)√(n² − m²)⁄n
(b)m⁄√(n² − m²)
(c)n⁄√(n² − m²)
(d)√(m² − n²)⁄n

Answer

Why

Correct — A. Build the right triangle from sin A = opposite ⁄ hypotenuse.

Opposite side = m, hypotenuse = n
Pythagoras: adjacent side = √(n² − m²)
cos A = adjacent ⁄ hypotenuse = √(n² − m²)⁄n

A lies between 0 and π⁄2, so cos A is positive and the positive root stands → option (a)

Why the others are wrong

(b)
m⁄√(n² − m²) — m⁄√(n² − m²) is tan A, opposite over adjacent. It divides by the adjacent side where cos A divides by the hypotenuse.
(c)
n⁄√(n² − m²) — n⁄√(n² − m²) is sec A, the reciprocal of cos A. Hypotenuse over adjacent is greater than 1 for an acute angle, while cos A is below 1.
(d)
√(m² − n²)⁄n — m² − n² is negative here: sin A = m⁄n lies between 0 and 1, so m² is less than n². Its square root is not a real number.

Concept

One trigonometric ratio fixes the others. Treat sin A = m⁄n as a right triangle with opposite side m and hypotenuse n, find the third side by Pythagoras, then read off the ratio asked.

The identity sin²A + cos²A = 1 gives the same result: cos²A = 1 − m²⁄n² = (n² − m²)⁄n². The interval (0, π⁄2) fixes the sign of the root.

π⁄2 radians is 90°, so A is acute and every trigonometric ratio of A is positive.

Key facts

sin A = opposite⁄hypotenuse, cos A = adjacent⁄hypotenuse, tan A = opposite⁄adjacent.
sin²A + cos²A = 1.
sec A = 1⁄cos A, so for an acute angle sec A is greater than 1.

Study next

1 + tan²A = sec²A and 1 + cot²A = cosec²ASigns of the trigonometric ratios in each quadrantWriting tan A in terms of sin A: sin A⁄√(1 − sin²A) for acute A

Common traps

Picking tan A, m⁄√(n² − m²), because it carries the same square root.
Writing the adjacent side as √(m² − n²), subtracting the squares in the wrong order.

The same m⁄n triangle is on 17 Sep 2025, 16:00, Quant Q.15, which asks for 1 + tan²A (keyed n²⁄(n² − m²)).

With sin A = x, 19 Sep 2025, 16:00, Quant Q.22 asks for cos²A (keyed 1 − x²) and 23 Sep 2025, 16:00, Quant Q.17 for tan A (keyed x⁄√(1 − x²)).

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 131 SSC CGL questions on it, every solved shift.