SSC CGL Heights and Distances Practice Test 5
15 SSC CGL heights and distances questions with chapter-wise explanations and exam-style options.
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SSC CGL Heights and Distances Practice Test 5
15 SSC CGL heights and distances questions with chapter-wise explanations and exam-style options.
Preview all 15 questions in SSC CGL Heights and Distances Practice Test 5 (no login required)
- Heights and Distances
1. The reciprocal of tan A is:
- A. sec A
- B. sin A
- C. cot A (Correct)
- D. cosec A
Explanation: cot A = 1 / tan A.
- Heights and Distances
2. The value of cos 45° is:
- A. 1/2
- B. 1/√2 (Correct)
- C. √3/2
- D. 1
Explanation: cos 45° = 1/√2.
- Heights and Distances
3. If sec A = 2, then tan² A equals:
- A. 1
- B. 2
- C. 3 (Correct)
- D. 4
Explanation: sec² A = 4, so tan² A = 3.
- Heights and Distances
4. If cot A = 7/24, then sin A is:
- A. 7/25
- B. 24/25 (Correct)
- C. 25/24
- D. 24/7
Explanation: Using a 7-24-25 triangle, sin A = 24/25.
- Heights and Distances
5. sin 12° is equal to:
- A. cos 78° (Correct)
- B. tan 78°
- C. cot 78°
- D. sec 12°
Explanation: sin θ = cos (90° - θ).
- Heights and Distances
6. A pole of height 6 m casts a shadow of length 6 m. Angle of elevation of the sun is:
- A. 30°
- B. 45° (Correct)
- C. 60°
- D. 75°
Explanation: tan θ = 6/6 = 1.
- Heights and Distances
7. The value of sec 45° is:
- A. 1/√2
- B. 1
- C. √2 (Correct)
- D. 2
Explanation: sec 45° = √2.
- Heights and Distances
8. If tan A = 4/3, then sin A + cos A is:
- A. 7/5 (Correct)
- B. 5/7
- C. 1
- D. 3/4
Explanation: With a 3-4-5 triangle, the sum is 4/5 + 3/5 = 7/5.
- Heights and Distances
9. Which is equal to cosec² A - 1?
- A. sec² A
- B. tan² A
- C. cot² A (Correct)
- D. sin² A
Explanation: cosec² A - 1 = cot² A.
- Heights and Distances
10. From a point on the ground, the angle of elevation of a tower top is 30°. Moving 20 m closer, the angle becomes 60°. Height of the tower is:
- A. 10√3 m (Correct)
- B. 15√3 m
- C. 20√3 m
- D. 30 m
Explanation: Let the nearer distance be x. Then h = x√3 and h = (x + 20)/√3. Solving gives x = 10 and h = 10√3 m.
- Heights and Distances
11. If sin A = 15/17, then sec A is:
- A. 17/8 (Correct)
- B. 8/17
- C. 17/15
- D. 15/8
Explanation: Using an 8-15-17 triangle, cos A = 8/17, so sec A = 17/8.
- Heights and Distances
12. The value of cot 60° is:
- A. 1/√3 (Correct)
- B. 1
- C. √3
- D. 2
Explanation: cot 60° = 1/√3.
- Heights and Distances
13. If sin A = 4/5, then sec A - tan A equals:
- A. 1/3 (Correct)
- B. 1/2
- C. 3/5
- D. 5/3
Explanation: With sin A = 4/5, cos A = 3/5, sec A = 5/3 and tan A = 4/3, so the difference is 1/3.
- Heights and Distances
14. The value of sin A / cosec A is:
- A. 1
- B. sin² A (Correct)
- C. cos² A
- D. tan A
Explanation: Since cosec A = 1 / sin A, the expression becomes sin² A.
- Heights and Distances
15. If tan A = 12/5, then cos A equals:
- A. 5/13 (Correct)
- B. 12/13
- C. 13/5
- D. 5/12
Explanation: Using a 5-12-13 triangle, cos A = 5/13.