Master Trigonometry: Practice and Sample Questions, Answers and Explanations for All Six Ratios

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.

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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.

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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

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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

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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

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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

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, 

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

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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

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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

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

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

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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

Answer 

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

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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

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°

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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

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