What Trigonometry Questions Test
- Trigonometric ratios and standard-angle values
- Identity-based simplification (sin²+cos²=1 family)
- Complementary-angle pairs (θ and 90°−θ)
- Compound and multiple angle formulas
- Heights and distances — angles of elevation/depression
Question bank coverage by sub-pattern
How Trigonometry Appears in SSC CPO
Trigonometry is part of the Quantitative Aptitude section (25% of the SSC CPO syllabus).
Trigonometry Solved Examples for SSC CPO
12 exam-pattern trigonometry questions from the practice bank, each with a full solution.
If cot θ = 8/15 and θ is an acute angle, what is the value of cos θ?
- A.8/17Correct
- B.15/17
- C.15/8
- D.8/15
Solution
Using a right triangle with adjacent=8, opposite=15, the hypotenuse = sqrt(8^2 + 15^2) = 17. cos θ = adjacent/hypotenuse = 8/17.
The angle of elevation of a kite from a point on the ground is 45°. If the length of the string between the kite and the point on the ground is 20 m, then what is the height of the kite?
- A.10√2 m
- B.10 m
- C.20√2 m
- D.20/√2 mCorrect
Solution
Let the height of the kite be 'h'. We are given the hypotenuse and need to find the perpendicular. So, we use the sine function. sin(45°) = Perpendicular/Hypotenuse = h/20. Since sin(45°) = 1/√2, we have 1/√2 = h/20. Therefore, h = 20/√2 m.
Simplify: 2sin(5x)cos(3x) + 2sin(6x)cos(2x)
- A.sin(8x) + sin(2x) + sin(8x) + sin(4x)Correct
- B.sin(8x) + sin(2x) + sin(4x)
- C.sin(2x) + sin(4x)
- D.2sin(8x) + sin(2x) + sin(4x)
Solution
Using the formula 2sinAcosB = sin(A+B) + sin(A-B). For 2sin(5x)cos(3x), A = 5x and B = 3x. So, 2sin(5x)cos(3x) = sin(5x+3x) + sin(5x-3x) = sin(8x) + sin(2x). For 2sin(6x)cos(2x), A = 6x and B = 2x. So, 2sin(6x)cos(2x) = sin(6x+2x) + sin(6x-2x) = sin(8x) + sin(4x). Therefore, 2sin(5x)cos(3x) + 2sin(6x)cos(2x) = sin(8x) + sin(2x) + sin(8x) + sin(4x).
What is the exact value of
- A.0
- B.2
- C.1Correct
- D.1/2
Solution
Since 70° and 20° are complementary angles (70° + 20° = 90°), cos 20° = cos(90° - 70°) = sin 70°. Therefore, sin 70° / cos 20° = sin 70° / sin 70° = 1.
Find the value of: (sin² 1° + sin² 3° + sin² 5° + ... + sin² 89°)
- A.22
- B.22.5Correct
- C.44
- D.45
Solution
We can pair sin²(1°) with sin²(89°), sin²(3°) with sin²(87°), and so on. Since sin(x) = cos(90° - x), we have sin²(x) + sin²(90° - x) = sin²(x) + cos²(x) = 1. There are 45 terms in the series. We can make 22 pairs adding up to 1 each. The middle term is sin²(45°) = (1/√2)² = 1/2. Therefore, the sum is 22 + 1/2 = 22.5.
What is the exact value of (sin 40° / cos 50°)?
- A.2
- B.0
- C.1Correct
- D.1/2
Solution
Since 40° and 50° are complementary angles (40° + 50° = 90°), cos 50° = cos(90° - 40°) = sin 40°. Therefore, sin 40° / cos 50° = sin 40° / sin 40° = 1.
What is the value of tan(32°) * tan(58°)?
- A.0
- B.1Correct
- C.2
- D.√3
Solution
Since 32° + 58° = 90°, tan(58°) = cot(32°). Thus, tan(32°) * tan(58°) = tan(32°) * cot(32°) = tan(32°) * (1/tan(32°)) = 1.
If cosec(75° + θ) - sec(15° - θ) - tan(55° + θ) + cot(35° - θ) = ?
- A.1
- B.0Correct
- C.-1
- D.1/2
Solution
cosec(75° + θ) = sec(90° - (75° + θ)) = sec(15° - θ). Also, tan(55° + θ) = cot(90° - (55° + θ)) = cot(35° - θ). Therefore, the expression becomes sec(15° - θ) - sec(15° - θ) - cot(35° - θ) + cot(35° - θ) = 0.
If sin(3A) = cos(A - 26°), where 3A is an acute angle, then what is the value of A?
- A.29°Correct
- B.30°
- C.28°
- D.26°
Solution
Since sin(3A) = cos(90° - 3A), we have cos(90° - 3A) = cos(A - 26°). Therefore, 90° - 3A = A - 26°. Solving for A, we get 4A = 116°, so A = 29°.
If tan(2θ) = cot(θ + 15°), where 2θ and θ + 15° are acute angles, then find the value of θ.
- A.25°Correct
- B.45°
- C.35°
- D.60°
Solution
Since tan(2θ) = cot(90° - 2θ), we have cot(90° - 2θ) = cot(θ + 15°). Therefore, 90° - 2θ = θ + 15°. Solving for θ, we get 3θ = 75°, so θ = 25°.
What is the exact value of (sin 35° / cos 55°)?
- A.1/2
- B.0
- C.2
- D.1Correct
Solution
Since 35° and 55° are complementary angles (35° + 55° = 90°), cos 55° = cos(90° - 35°) = sin 35°. Therefore, sin 35° / cos 55° = sin 35° / sin 35° = 1.
What is the exact value of (sin 25° / cos 65°)?
- A.0
- B.1/2
- C.2
- D.1Correct
Solution
Since 25° and 65° are complementary angles (25° + 65° = 90°), cos 65° = cos(90° - 25°) = sin 25°. Therefore, sin 25° / cos 65° = sin 25° / sin 25° = 1.
Trigonometry — Frequently Asked Questions
How do I approach a heights-and-distances question?
Draw the right triangle first and mark the known angle and side. The question is then a single ratio — tan for height vs base, sin/cos when the hypotenuse is involved.
Do I need to memorise every trig formula?
No. Standard-angle values, the three Pythagorean identities and complementary-angle relations cover the bulk of exam-pattern questions; compound-angle formulas cover most of the rest.
How many Trigonometry practice questions does Parikshala have for SSC CPO?
130 exam-pattern Trigonometry questions for SSC CPO Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.