SSC CGL 2024 Tier-I · 24 September 2024, Shift 2 · Quantitative Aptitude · Question 25 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.

If 3 tan A = 4 and A is an acute angle, then the value of 4sinA + 3cosA is:

(a)1
(b)4
(c)5
(d)3

Answer

Why

Correct — C. The image gives 3 tanA = 4 with A acute, and asks for 4 sinA + 3 cosA.

tanA = 4⁄3, so build the right triangle:
opposite = 4, adjacent = 3
hypotenuse = √(4² + 3²) = √25 = 5

A is acute, so both ratios are positive:
sinA = 4⁄5
cosA = 3⁄5

Substitute:
4 sinA + 3 cosA = 4 × 4⁄5 + 3 × 3⁄5
= 16⁄5 + 9⁄5 = 25⁄5 = 5 → option (c)

Why the others are wrong

(a)
1 — With A acute both ratios are positive, so the total is larger than either piece. 4 sinA on its own is 16⁄5 = 3.2, and 3 cosA adds another 1.8 to it, so 1 is ruled out before the arithmetic finishes.
(b)
4 — Reading the ratio backwards as tanA = 3⁄4 gives sinA = 3⁄5 and cosA = 4⁄5, and then 4 sinA + 3 cosA = 24⁄5 = 4.8 — still not 4. The stem says 3 tanA = 4, so tanA is 4⁄3.
(d)
3 — 3 is below 4 sinA = 3.2 by itself, so it cannot survive the addition of a positive 3 cosA = 1.8. Once the hypotenuse 5 is found the total is fixed at 25⁄5.

Concept

One ratio fixes all six. Given tanA = 4⁄3 with A acute, the 3-4-5 right triangle supplies sinA = 4⁄5 and cosA = 3⁄5 directly, with no identity work at all.

The word acute is doing real work here. It places A in the first quadrant, where sine and cosine are both positive, so the signs are settled and there is no second solution to weigh.

The coefficients are chosen to match the sides: 4 sinA + 3 cosA becomes (4×4 + 3×3)⁄5 = 25⁄5, which is why a messy-looking expression lands on a whole number.

The pattern generalises and is worth carrying. If tanA = p⁄q with A acute, then p sinA + q cosA = √(p² + q²) — the hypotenuse itself. Here p = 4 and q = 3, so the value is 5 without substituting anything.

Key facts

3 tanA = 4 gives tanA = 4⁄3, so the right triangle has legs 3 and 4 and hypotenuse 5.
For A acute, sinA = 4⁄5 and cosA = 3⁄5, both positive.
4 sinA + 3 cosA = 16⁄5 + 9⁄5 = 25⁄5 = 5.
If tanA = p⁄q with A acute, then p sinA + q cosA equals √(p² + q²), the hypotenuse.

Study next

Recovering all six ratios from one given ratio using a right trianglePythagorean triples that recur in SSC trigonometry, such as 3-4-5 and 5-12-13Why the acute condition removes the sign ambiguity in sine and cosine

Common traps

Using 4 and 3 as sine and cosine without dividing by the hypotenuse.
Inverting the given ratio and taking tanA as 3⁄4.
Matching the coefficients to the wrong ratios, when 4 pairs with sinA here and 3 with cosA.

Whenever a single ratio is given and a combination of sine and cosine is asked for, the work runs the same way: build the triangle, find the hypotenuse, substitute.

Trigonometry in this paper is also set as pure identity work — Quant Q.19 asks for tan²A + cot²A given tanA + cotA = 2, and Quant Q.20 expands (sinA + cosecA)² + (cosA + secA)².

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