Trigonometry

Maths, SSC-Punjab Class 9. This chapter is split into 46 topics and has 1,062 practice questions. The topics covered are listed below, followed by 10 free sample questions from the chapter.

Topics in this chapter

1. Review Exercise 6 - Conversions, Identities and Applications 22 questions
2. Evaluating Expressions with Special Angles - Exercise 6.4 30 questions
3. Finding Angles from Ratios (Inverse Trigonometric Ratios) 26 questions
4. Solving Right Triangles - Examples 15 to 18 22 questions
5. Isosceles Triangle, Shadow and Kite Problems 21 questions
6. Converting Decimal Degrees to Degrees, Minutes and Seconds 23 questions
7. Proving Identities - Exercise 6.3 Q2 to Q12 21 questions
8. Positive and Negative Angles and Quadrants 24 questions
9. Angle of Elevation and Angle of Depression - Definitions 21 questions
10. Angle of Depression Problems - Example 20 and Exercise 6.6 22 questions
11. Reciprocal and Quotient Relations of Trigonometric Ratios 23 questions
12. Converting Degrees to Radians 30 questions
13. Special Angle Values in Radians and Mixed Applications 23 questions
14. Trigonometric Ratios of 30 and 60 Degrees (pi/6 and pi/3) 25 questions
15. Solving Right Triangles - Exercise 6.5 Style Problems 29 questions
16. Identities tan^2 + 1 = sec^2 and 1 + cot^2 = cosec^2 - Applications 23 questions
17. Valid and Invalid Trigonometric Identities 23 questions
18. Degree Measure - Minutes and Seconds 21 questions
19. Simplifying Trigonometric Expressions with Identities 24 questions
20. Ratios in the Triangles of Exercise 6.2 25 questions
21. Proving Identities - Worked Examples 10 to 13 21 questions
22. Review Exercise 6 - MCQs 22 questions
23. Converting Degrees, Minutes and Seconds to Decimal Degrees 22 questions
24. History of Trigonometry and Direct versus Indirect Measurement 20 questions
25. Two-Observation Elevation Problems - Exercise 6.6 Q6 and Q8 20 questions
26. Trigonometric Ratios of Complementary Angles 22 questions
27. Radian Measure - Definition and Facts 24 questions
28. Area of a Sector 20 questions
29. Finding the Angle or Radius from Arc Length or Area 22 questions
30. Sector Problems - Percentage of Area and Cone from a Sector 21 questions
31. Arc Length of a Sector 20 questions
32. Computing Trigonometric Ratios from Right Triangles 25 questions
33. Turns, Radians and Degrees Table 21 questions
34. Trigonometric Ratios of 45 Degrees (pi/4) 23 questions
35. Pythagoras Theorem Applications - Exercise 6.5 Q3, Q5 to Q9 21 questions
36. Angle of Elevation Problems - Example 19 and Exercise 6.6 23 questions
37. Angles in Standard Position and Types of Angles 25 questions
38. Solution of a Right Triangle - Elements and Cases 22 questions
39. Pythagoras Theorem and the Fundamental Identities 22 questions
40. The Identity sin^2 + cos^2 = 1 - Numerical and Algebraic Use 24 questions
41. Co-terminal Angles 21 questions
42. Finding the Remaining Ratios from One Ratio 30 questions
43. Table of Trigonometric Ratios of Special Angles 24 questions
44. Complementary Angle Relations - Exercise 6.2 Q4 and Q5 22 questions
45. Trigonometric Ratios - Definitions and Triangle Labels 24 questions
46. Converting Radians to Degrees 23 questions

Sample questions: Trigonometry

10 sample questions with the correct answer and an explanation. Try to answer each one before reading the explanation.

Q1. Convert 123.456° into degrees, minutes and seconds.

• A. 123° 46′ 0″
• B. 123° 37′ 21.6″
• C. 123° 27′ 21.6″(correct answer)
• D. 123° 27′ 34″

Answer: C.Degrees: 123. Minutes: 0.456 × 60 = 27.3600, so 27′. Seconds: the decimal part of the minutes × 60 = 21.6″. Answer: 123° 27′ 21.6″.

Q2. Convert 58.7891° into degrees, minutes and seconds.

• A. 58° 57′ 20.76″
• B. 58° 47′ 20.76″(correct answer)
• C. 58° 48′ 20.76″
• D. 58° 47′ 33″

Answer: B.Degrees: 58. Minutes: 0.7891 × 60 = 47.3460, so 47′. Seconds: the decimal part of the minutes × 60 = 20.76″. Answer: 58° 47′ 20.76″.

Q3. Convert 90.5678° into degrees, minutes and seconds.

• A. 90° 34′ 16″
• B. 90° 57′ 0″
• C. 90° 34′ 4.08″(correct answer)
• D. 90° 44′ 4.08″

Answer: C.Degrees: 90. Minutes: 0.5678 × 60 = 34.0680, so 34′. Seconds: the decimal part of the minutes × 60 = 4.08″. Answer: 90° 34′ 4.08″.

Q4. Who among the following was among the first to replace chord functions with the modern sine function?

• A. Al-Battani(correct answer)
• B. Newton
• C. Hipparchus
• D. Leibniz

Answer: A.The box says Al-Battani (c. 858-929 CE) replaced chord functions with the sine function and calculated tables of sines and tangents.

Q5. From the top of a 50 m tower the angle of depression of a car on the ground is 30°. How far is the car from the foot of the tower?

• A. 25.00 m
• B. 86.60 m(correct answer)
• C. 28.87 m
• D. 100.00 m

Answer: B.The angle of elevation of the observer from the object equals the angle of depression (30°), so distance = 50/tan 30° = 86.60 m.

Q6. An observer on a 120 m lighthouse sees a ship at an angle of depression of 60°. The ship's distance from the lighthouse is:

• A. 207.85 m
• B. 69.28 m(correct answer)
• C. 138.56 m
• D. 103.92 m

Answer: B.The angle of elevation of the observer from the object equals the angle of depression (60°), so distance = 120/tan 60° = 69.28 m.

Q7. From the top of a 90 m tower the angle of depression of a car on the ground is 30°. How far is the car from the foot of the tower?

• A. 51.96 m
• B. 45.00 m
• C. 180.00 m
• D. 155.88 m(correct answer)

Answer: D.The angle of elevation of the observer from the object equals the angle of depression (30°), so distance = 90/tan 30° = 155.88 m.

Q8. An observer on a 200 m lighthouse sees a ship at an angle of depression of 60°. The ship's distance from the lighthouse is:

• A. 346.41 m
• B. 230.94 m
• C. 173.21 m
• D. 115.47 m(correct answer)

Answer: D.The angle of elevation of the observer from the object equals the angle of depression (60°), so distance = 200/tan 60° = 115.47 m.

Q9. The angle of depression of a boat from the top of a 100 m cliff is 45°. The horizontal distance of the boat from the cliff is:

• A. 70.71 m
• B. 141.42 m
• C. 100.00 m(correct answer)
• D. 200.00 m

Answer: C.The angle of elevation of the observer from the object equals the angle of depression (45°), so distance = 100/tan 45° = 100.00 m.

Q10. From the top of a 150 m tower the angle of depression of a car on the ground is 30°. How far is the car from the foot of the tower?

• A. 259.81 m(correct answer)
• B. 86.60 m
• C. 300.00 m
• D. 75.00 m

Answer: A.The angle of elevation of the observer from the object equals the angle of depression (30°), so distance = 150/tan 30° = 259.81 m.