SSC CGL Heights and Distances Practice Test 3
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 3
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 3 (no login required)
- Heights and Distances
1. If cos A = 3/5, then sin A is:
- A. 4/5 (Correct)
- B. 3/4
- C. 5/4
- D. 2/5
Explanation: Using a 3-4-5 triangle, sin A = 4/5.
- Heights and Distances
2. If tan A = 1/√3, then A is:
- A. 30° (Correct)
- B. 45°
- C. 60°
- D. 90°
Explanation: tan 30° = 1/√3.
- Heights and Distances
3. The value of cosec 90° is:
- A. 0
- B. 1 (Correct)
- C. 2
- D. Undefined
Explanation: cosec 90° = 1.
- Heights and Distances
4. The value of (sin A + cos A)² is:
- A. 1 + 2sin A cos A (Correct)
- B. 1 - 2sin A cos A
- C. sin² A + cos² A
- D. 2
Explanation: Expand the square and use sin² A + cos² A = 1.
- Heights and Distances
5. A 10 m pole casts a shadow 10√3 m long. The angle of elevation of the sun is:
- A. 30° (Correct)
- B. 45°
- C. 60°
- D. 75°
Explanation: tan θ = 10 / 10√3 = 1/√3, so θ = 30°.
- Heights and Distances
6. If sec A = 13/12, then tan A is:
- A. 5/12 (Correct)
- B. 12/5
- C. 13/5
- D. 5/13
Explanation: With sec A = 13/12, use a 5-12-13 triangle.
- Heights and Distances
7. The value of sin 60° is:
- A. 1/2
- B. 1/√2
- C. √3/2 (Correct)
- D. 1
Explanation: sin 60° = √3/2.
- Heights and Distances
8. cot 18° is equal to:
- A. tan 72° (Correct)
- B. cos 72°
- C. sin 72°
- D. sec 18°
Explanation: cot θ = tan (90° - θ).
- Heights and Distances
9. If cos A = 4/5, then sec A + tan A equals:
- A. 7/5
- B. 9/5
- C. 2 (Correct)
- D. 3
Explanation: sec A = 5/4 and tan A = 3/4, so sum = 2.
- Heights and Distances
10. From a point 15 m from the base of a building, the angle of elevation of its top is 60°. Height of the building is:
- A. 5√3 m
- B. 10√3 m
- C. 15√3 m (Correct)
- D. 30 m
Explanation: tan 60° = height / 15, so height = 15√3 m.
- Heights and Distances
11. If sin A = 3/5, then (1 - cos A)(1 + cos A) equals:
- A. 3/5
- B. 9/25 (Correct)
- C. 16/25
- D. 1
Explanation: This equals 1 - cos² A = sin² A = 9/25.
- Heights and Distances
12. The value of cot 30° is:
- A. 1/√3
- B. 1
- C. √3 (Correct)
- D. 2
Explanation: cot 30° = √3.
- Heights and Distances
13. If tan A = 8/15, then sec A is:
- A. 15/17
- B. 17/15 (Correct)
- C. 8/17
- D. 15/8
Explanation: Using an 8-15-17 triangle, sec A = 17/15.
- Heights and Distances
14. The value of cos² A + sin² A + tan² A - sec² A is:
- A. -1
- B. 0 (Correct)
- C. 1
- D. tan² A
Explanation: Use cos² A + sin² A = 1 and sec² A - tan² A = 1.
- Heights and Distances
15. sin 72° is equal to:
- A. cos 18° (Correct)
- B. tan 18°
- C. sec 18°
- D. cot 72°
Explanation: sin θ = cos (90° - θ).