SSC CGL 2025 Tier-I · 17 September 2025, Shift 1 · Quantitative Aptitude · Question 14 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 13 September 2026.

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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 θ = 12⁄13, and θ ∈ (0, π/2), then what is the value of tan θ?

(a)9⁄16
(b)25⁄9
(c)12⁄5
(d)16⁄9

Answer

Why

Correct — C.
sin θ = opposite ⁄ hypotenuse = 12⁄13

Adjacent side = √(13² − 12²) = √(169 − 144) = √25 = 5
θ lies in (0, π⁄2), so cos θ = +5⁄13

tan θ = sin θ ⁄ cos θ = (12⁄13) ÷ (5⁄13) = 12⁄5 → option (c)

Why the others are wrong

(a)
9⁄16 — 9⁄16 is less than 1, but here sin θ = 12⁄13 is larger than cos θ = 5⁄13, so tan θ must be greater than 1.
(b)
25⁄9 — 25⁄9 ≈ 2.78, not 12⁄5 = 2.4. The sides here are 5, 12 and 13, and no ratio of two of them equals 25⁄9.
(d)
16⁄9 — 16⁄9 ≈ 1.78, not 2.4. Tangent is opposite ÷ adjacent, 12 ÷ 5 here, and 16⁄9 is not a ratio of any two of 5, 12 and 13.

Concept

Given one trigonometric ratio, draw the right triangle. sin θ = 12⁄13 fixes the opposite side at 12 and the hypotenuse at 13, and Pythagoras gives the third side.

5, 12, 13 is a Pythagorean triple (25 + 144 = 169), so the adjacent side is 5 and every other ratio follows: cos θ = 5⁄13, tan θ = 12⁄5.

θ ∈ (0, π⁄2) puts the angle in the first quadrant, where sin, cos and tan are all positive. Without that condition cos θ could be −5⁄13, and tan θ would then be −12⁄5.

Key facts

5, 12, 13 is a Pythagorean triple: 5² + 12² = 25 + 144 = 169 = 13².
tan θ = sin θ ⁄ cos θ = opposite ⁄ adjacent.
In the first quadrant, θ ∈ (0, π⁄2), sin, cos and tan are all positive.

Study next

All six ratios from one given ratio, using the triples 3-4-5, 8-15-17 and 7-24-25Signs of the trigonometric ratios in each quadranttan θ in terms of sin θ: sin θ ⁄ √(1 − sin²θ)

Common traps

Putting the adjacent side on top and writing tan θ = 5⁄12: tangent is opposite over adjacent, 12⁄5.
Taking cos θ = 1 − sin θ = 1⁄13 instead of √(1 − sin²θ) = 5⁄13.

19 Sep 2025, 09:00, Quant Q.22 starts the same way from sin x = 3⁄5 in (0, π⁄2): tan x = 3⁄4, so (1 + tan x) ⁄ (1 − tan x) = 7 (option b).

18 Sep 2025, 12:30, Quant Q.14 uses the same 5-12-13 triangle, turning cos B = 12⁄13 into sin B = 5⁄13 on the way to sin(A + B) = 56⁄65 (option a).

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.