Live Practice Test

SSC CGL Heights and Distances Practice Test 2

15 SSC CGL heights and distances questions with chapter-wise explanations and exam-style options.

Timer-based practiceDetailed answer reviewMobile-friendly flow
SSC CGL Sponsor
High Intent Competitive Exam Traffic

Leaderboard slot above the practice session for revision tools, test series, books, and career products.

SSC CGL Practice Session

SSC CGL Heights and Distances Practice Test 2

15 SSC CGL heights and distances questions with chapter-wise explanations and exam-style options.

Questions
15
Marking
+2 correct, -0.5 wrong
Format
15-question ad-supported session
Preview all 15 questions in SSC CGL Heights and Distances Practice Test 2 (no login required)
  1. Heights and Distances

    1. The value of cot 45° is:

    • A. 0
    • B. 1 (Correct)
    • C. √3
    • D. Undefined

    Explanation: cot 45° = 1.

  2. Heights and Distances

    2. sec² θ - tan² θ is:

    • A. 0
    • B. 1 (Correct)
    • C. sin² θ
    • D. cos² θ

    Explanation: From 1 + tan² θ = sec² θ, we get sec² θ - tan² θ = 1.

  3. Heights and Distances

    3. The value of sin 45° is:

    • A. 1/2
    • B. 1/√2 (Correct)
    • C. √3/2
    • D. 1

    Explanation: sin 45° = 1/√2.

  4. Heights and Distances

    4. tan 25° is equal to:

    • A. cot 25°
    • B. cot 65° (Correct)
    • C. sec 65°
    • D. sin 65°

    Explanation: tan θ = cot (90° - θ).

  5. Heights and Distances

    5. If tan A = 5/12, then sin A is:

    • A. 5/13 (Correct)
    • B. 12/13
    • C. 13/5
    • D. 5/12

    Explanation: Using a 5-12-13 triangle, sin A = 5/13.

  6. Heights and Distances

    6. The value of cos 30° is:

    • A. 1/2
    • B. √3/2 (Correct)
    • C. 1/√3
    • D. 1

    Explanation: cos 30° = √3/2.

  7. Heights and Distances

    7. cosec² A - cot² A is:

    • A. 0
    • B. 1 (Correct)
    • C. sec² A
    • D. sin² A

    Explanation: From 1 + cot² A = cosec² A.

  8. Heights and Distances

    8. If a ladder makes an angle of 60° with the ground and its length is 10 m, the height reached is:

    • A. 5 m
    • B. 5√3 m (Correct)
    • C. 10/√3 m
    • D. 10 m

    Explanation: Height = 10 × sin 60° = 5√3 m.

  9. Heights and Distances

    9. The value of tan 60° is:

    • A. 1/√3
    • B. 1
    • C. √3 (Correct)
    • D. 2

    Explanation: tan 60° = √3.

  10. Heights and Distances

    10. If sin A = 8/17, then cot A is:

    • A. 8/15
    • B. 15/8 (Correct)
    • C. 17/8
    • D. 15/17

    Explanation: Using an 8-15-17 triangle, cot A = 15/8.

  11. Heights and Distances

    11. If sin A = 1/2 and A is acute, then cos A equals:

    • A. √3/2 (Correct)
    • B. 1/2
    • C. 1/√2
    • D. 2/√3

    Explanation: For an acute angle with sin A = 1/2, cos A = √3/2.

  12. Heights and Distances

    12. sec 35° is equal to:

    • A. cosec 55° (Correct)
    • B. sin 55°
    • C. tan 55°
    • D. cot 35°

    Explanation: sec θ = cosec (90° - θ).

  13. Heights and Distances

    13. The value of cos 90° is:

    • A. 0 (Correct)
    • B. 1/2
    • C. 1
    • D. Undefined

    Explanation: cos 90° = 0.

  14. Heights and Distances

    14. The angle of elevation of the top of a tower from a point 20 m away is 45°. The height of the tower is:

    • A. 10 m
    • B. 20 m (Correct)
    • C. 20√2 m
    • D. 40 m

    Explanation: tan 45° = height / 20 = 1, so height = 20 m.

  15. Heights and Distances

    15. If tan A = 3/4, then sin A × cos A equals:

    • A. 3/25
    • B. 12/25 (Correct)
    • C. 7/25
    • D. 24/25

    Explanation: sin A = 3/5 and cos A = 4/5, so product = 12/25.