Practice Questions with Answers to Learn All Six Trigonometric Ratios
Mastering trigonometry starts with understanding all six ratios: sin, cos, tan, csc, sec, cot.
Here is a question set to practice and master this chapter. Each question includes answers and explanations.
Part A: Concept Questions
(10 Practice Questions)
Q1: What is the formula for sin θ in a right triangle
Answer: sin θ = Opposite / Hypotenuse
Explanation: Sine is the ratio of the opposite side to the hypotenuse.
Q2: What is the formula for cos θ?
Answer: cos θ = Adjacent / Hypotenuse
Q3: What is the formula for tan θ?
Answer: tan θ = Opposite / Adjacent
Q4: What is the reciprocal of sin θ?
Answer: csc θ (cosecant) = 1/sin θ
Q5: What is sec θ in terms of sides?
Answer: sec θ = Hypotenuse / Adjacent
Q6: What is cot θ in terms of sides?
Answer: cot θ = Adjacent / Opposite
Q7: Name the three primary trigonometric ratios.
Answer: sin θ, cos θ, tan θ
Q8: Name the three reciprocal trigonometric ratios.
Answer: csc θ, sec θ, cot θ
Q9: Which trigonometric ratios are always less than 1 in a right triangle?
Answer: sin θ and cos θ
Explanation: Opposite and adjacent sides are always shorter than the hypotenuse.
Q10: Which trigonometric ratio represents the slope of a line in a triangle?
Answer: tan θ
Explanation: Tan θ = height/base = slope.
Part B: Fill in the Blanks (10 Practice Questions)
Q11. sin θ = ___ / ___→
Answer: Opposite / Hypotenuse
Q12. cos θ = ___ / ___
→
Answer: Adjacent / Hypotenuse
Q13. tan θ = ___ / ___
→
Answer: Opposite / Adjacent
Q14. csc θ = ___ / ___
→
Answer: Hypotenuse / Opposite
Q15. sec θ = ___ / ___
→
Answer: Hypotenuse / Adjacent
Q16. cot θ = ___ / ___
→
Answer: Adjacent / Opposite
Q17. Reciprocal of tan θ is ___
→
Answer: cot θ
Q18. Reciprocal of cos θ is ___
→
Answer: sec θ
Q19. Reciprocal of sin θ is ___
→
Answer: csc θ
Q20. sin 30° = ___
→
Answer: 1/2
Part C: True/False (10 Practice Questions)
Q21. sin θ = Hypotenuse / Opposite→ False
Q22. cos θ is the reciprocal of sec θ
→ True
Q23. tan θ = sin θ / cos θ
→ True
Q24. cot θ = sin θ / cos θ
→ False
Q25. sec θ is always ≥ 1
→ True
Q26. sin 90° = 0
→ False
Q27. cos 0° = 1
→ True
Q28. tan 45° = 1
→ True
Q29. cot 45° = 1
→ True
Q30. sec 90° = 0
→ False
Part D: Direct Computation (10 Practice Questions)
Q31. Find sin 30° →Answer: 1/2
Q32. Find cos 30° →
Answer: √3/2
Q33. Find tan 30° →
Answer: 1/√3
Q34. Find csc 30° →
Answer: 2
Q35. Find sec 30° →
Answer: 2/√3
Q36. Find cot 30° →
Answer: √3
Q37. Find sin 45° →
Answer: √2/2
Q38. Find cos 45° →
Answer: √2/2
Q39. Find tan 45° →
Answer: 1
Q40. Find sec 45° →
Answer: √2
Part E: Real-Life Word Problems (10 Practice Questions)
Q41. A ladder 10 m long reaches a wall 8 m high. Find sin θ, cos θ, and tan θ.
Answer:
sin θ = 8/10 = 0.8
cos θ = 6/10 = 0.6
tan θ = 8/6 = 4/3
cos θ = 6/10 = 0.6
tan θ = 8/6 = 4/3
Q42. A tree 15 m high casts a 20 m shadow. Find tan θ.
Answer: tan θ = 15/20 = 3/4
Q43. A flagpole 12 m tall casts a 5 m shadow. Find the angle of elevation.
Answer: tan θ = 12/5 → θ ≈ 67.38°
Q44. A ladder 5 m long touches a wall at 3 m height. Find sin θ.
Answer: sin θ = 3/5 = 0.6
Q45. A ramp 10 m long rises 6 m. Find cos θ.
Answer: cos θ = 8/10 = 0.8
Q46. A kite string is 13 m and height is 12 m. Find all six ratios.
Answer: O=12, H=13, A=5
sin=12/13,
sin=12/13,
cos=5/13,
tan=12/5,
csc=13/12,
sec=13/5,
cot=5/12
Q47. A bridge cable of 25 m supports a pole of 7 m. Find sin θ.
Answer: sin θ = 7/25 = 0.28
Q48. A slope rises 9 m for every 12 m base. Find tan θ.
Answer: tan θ = 9/12 = 3/4
Q49. A ladder forms 60° with ground. Hypotenuse = 10 m. Find height.
Answer: h = 10 × sin 60° = 10×(√3/2) ≈ 8.66 m
Q50. A light pole casts a 24 m shadow at 16° elevation. Find height.
Answer: h = tan 16° × 24 ≈ 6.87 m
Part F: Conceptual Questions (50)
Q1: Define sin θ in a right triangle.Answer: sin θ = Opposite / Hypotenuse
Explanation: Sine is the ratio of the side opposite the angle to the hypotenuse.
Q2: Define cos θ.
Answer: cos θ = Adjacent / Hypotenuse
Q3: Define tan θ.
Answer: tan θ = Opposite / Adjacent
Q4: Define csc θ.
Answer: csc θ = Hypotenuse / Opposite
Explanation: It is the reciprocal of sin θ.
Q5: Define sec θ.
Answer: sec θ = Hypotenuse / Adjacent
Explanation: It is the reciprocal of cos θ.
Q6: Define cot θ.
Answer: cot θ = Adjacent / Opposite
Explanation: It is the reciprocal of tan θ.
Q7: Which trigonometric ratios are primary?
Answer: sin θ, cos θ, tan θ
Q8: Which are reciprocal ratios?
Answer: csc θ, sec θ, cot θ
Q9: Which ratio shows slope in a right triangle?
Answer: tan θ
Q10: Which ratio is undefined for θ = 90°?
Answer: tan θ and sec θ
Q11: If sin θ = 0, what is θ?
Answer: 0° or 180°
Explanation: Sine is 0 when the opposite side is 0.
Q12: If cos θ = 0, what is θ?
Answer: 90° or 270°
Q13: Which ratio is Opposite/Adjacent?
Answer: tan θ
Q14: Which ratio is Adjacent/Opposite?
Answer: cot θ
Q15: Name the 6 trigonometric ratios.
Answer: sin, cos, tan, csc, sec, cot
Part G: True or False
Decide if each statement is True or False. Explanations are provided.
1. sin θ = Hypotenuse / Opposite→ False
Explanation: sin θ = Opposite / Hypotenuse
2. cos θ is the reciprocal of sec θ
→ True
3. tan θ = sin θ / cos θ
→ True
4. cot θ = sin θ / cos θ
→ False
Explanation: cot θ = cos θ / sin θ
Explanation: cot θ = cos θ / sin θ
5. sec θ is always ≥ 1
→ True
6. sin 90° = 0
→ False
7. cos 0° = 1
→ True
8. tan 45° = 1
→ True
9. cot 45° = 1
→ True
10. sec 90° = 0
→ False
Explanation: sec 90° = 1/cos 90° = 1/0 → undefined
Explanation: sec 90° = 1/cos 90° = 1/0 → undefined
11. csc 30° = 2
→ True
12. tan 30° = √3
→ False (tan 30° = 1/√3)
13. sec 60° = 2
→ True
14. cot 60° = 1/√3
→ True
15. sin²θ + cos²θ = 1 for all θ
→ True
Part H: Match the Following
Match trigonometric ratios with their correct definitions.
Column 1 (Ratios) | Column 2 (Meanings)
- 1. sin θ
- 2. cos θ
- 3. tan θ
- 4. csc θ
- 5. sec θ
- 6. cot θ
- a. Hypotenuse / Opposite
- b. Hypotenuse / Adjacent
- c. Opposite / Hypotenuse
- d. Adjacent / Hypotenuse
- e. Opposite / Adjacent
- f. Adjacent / Opposite
1. sin θ → c.
Opposite / Hypotenuse
2. cos θ → d.
Adjacent / Hypotenuse
3. tan θ → e.
Opposite / Adjacent
4. csc θ → a.
Hypotenuse / Opposite
5. sec θ → b.
Hypotenuse / Adjacent
6. cot θ → f.
Adjacent / Opposite
Part I: Real-Life Word Problems
Apply trigonometric ratios to real-life situations like ladders, trees, ramps, and towers.
1. A ladder 10 m long reaches a wall 8 m high. Find sin θ, cos θ, and tan θ.Answer: sin θ = 8/10 = 0.8, cos θ = 6/10 = 0.6, tan θ = 8/6 = 4/3
Explanation: This is a 6-8-10 right triangle (Pythagoras triple).
2. A tree 15 m high casts a 20 m shadow. Find the angle of elevation of the sun.
Answer: tan θ = 15/20 = 3/4 → θ ≈ 36.87°
Explanation: Tan = Opposite / Adjacent.
3. A 12 m flagpole casts a 5 m shadow. Find the angle of elevation.
Answer: tan θ = 12/5 = 2.4 → θ ≈ 67.38°
4. A ramp 10 m long rises 6 m. Find sin θ.
Answer: sin θ = 6/10 = 3/5
5. A kite string 13 m long reaches a height of 12 m. Find all six trigonometric ratios.
Answer: O = 12, H = 13, A = 5
sin = 12/13, cos = 5/13, tan = 12/5, csc = 13/12, sec = 13/5, cot = 5/12
sin = 12/13, cos = 5/13, tan = 12/5, csc = 13/12, sec = 13/5, cot = 5/12
6. A bridge cable 25 m supports a pole 7 m high. Find sin θ.
Answer: sin θ = 7/25 = 0.28
7. A slope rises 9 m for every 12 m base. Find tan θ.
Answer: tan θ = 9/12 = 3/4
8. A ladder forms 60° with the ground. Its length is 10 m. Find the height it reaches.
Answer: h = 10 * sin 60° = 10 * √3/2 ≈ 8.66 m
9. A light pole casts a 24 m shadow at 16° elevation. Find its height.
Answer: h = tan 16° * 24 ≈ 6.87 m
10. A drone flies 50 m high and is 120 m away horizontally from the observer. Find the angle of elevation.
Answer: tan θ = 50/120 = 5/12 → θ ≈ 22.62°
Part J: Mixed Problem-Solving
Combine conceptual knowledge, computations, and word problems for mastery.
Q1. If sin θ = 3/5, find cos θ and tan θ.Answer: cos θ = 4/5, tan θ = 3/4
Explanation: Pythagoras → 5² - 3² = 4²
Q2. If tan θ = 5/12, find all six trigonometric ratios.
Answer: sin = 5/13, cos = 12/13, tan = 5/12, csc = 13/5, sec = 13/12, cot = 12/5
Q3. A 20 m tall building casts a 15 m shadow. Find sin θ and cos θ.
Answer: tan θ = 20/15 = 4/3 → θ ≈ 53.13°
sin θ ≈ 0.799, cos θ ≈ 0.601
Q4. If sec θ = 5/4, find tan θ.
Answer: tan²θ = sec²θ - 1 = (25/16) - 1 = 9/16 → tan θ = 3/4
Q5. If csc θ = 13/5, find cos θ.
Answer: sin θ = 5/13 → cos²θ = 1 - (25/169) = 144/169 → cos θ = 12/13
Q6. A tower is 50 m high. A point on the ground is 30 m from the base. Find all six ratios.
Answer: sin = 5/√34, cos = 3/√34, tan = 5/3, csc = √34/5, sec = √34/3, cot = 3/5
Q7. A hill has an angle of elevation of 25° from 100 m away. Find its height.
Answer: h = tan 25° * 100 ≈ 46.63 m
Q8. If cot θ = 7/24, find all six ratios.
Answer: tan = 24/7 → Hypotenuse = 25
sin = 24/25, cos = 7/25, tan = 24/7, csc = 25/24, sec = 25/7, cot = 7/24
sin = 24/25, cos = 7/25, tan = 24/7, csc = 25/24, sec = 25/7, cot = 7/24
Q9. Verify: sin²30° + cos²30° = 1.
Answer: sin² = 1/4, cos² = 3/4 → sum = 1
Q10. A flagpole casts a 10 m shadow at 60° elevation. Find the pole height.
Answer: h = tan 60° * 10 ≈ 17.32 m
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