CBSE Class 10 Mathematics Chapter 9 Some Applications of Trigonometry Worksheet with Answers

Chapter 9 applies trigonometric ratios learned in Chapter 8 to solve real-world problems involving heights and distances. The core concepts are angle of elevation (angle formed with horizontal when looking upward at an object) and angle of depression (angle formed with horizontal when looking downward). Problems typically involve towers, buildings, cliffs, ships, aeroplanes and poles. You use right-triangle properties with tan θ = perpendicular/base, sin θ = perpendicular/hypotenuse, and cos θ = base/hypotenuse. The chapter emphasises drawing accurate diagrams, identifying the right triangle, choosing the appropriate trigonometric ratio, and solving for unknown heights or distances. Key values to memorise: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, sin 30° = 1/2, cos 30° = √3/2, sin 45° = cos 45° = 1/√2, sin 60° = √3/2, cos 60° = 1/2.

This worksheet is designed for a 90-minute timed practice session under exam-like conditions. Difficulty level: Moderate to Advanced, suitable for students aiming for 85+ marks in CBSE Class 10 Mathematics board exams. The question distribution mirrors the CBSE blueprint: Section A tests quick conceptual clarity through MCQs; Section B assesses formula recall; Section C checks understanding of definitions; Section D contains 2-3 mark application problems; Section E has 5-mark HOTS and case-study questions. Use standard trigonometric tables or values. Calculators are not required. Show all working clearly for numerical problems. Diagrams must be neat and labelled. After completing all sections, refer to the answer key at the end for self-assessment. Mark your errors, understand the solution approach, and retry incorrect questions after 24 hours for effective revision.

Choose the correct option for each question. Each MCQ carries 1 mark. No negative marking. Time allocation: 10 minutes for this section. Q1. A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side to reach a window 12 m high. The width of the street is: (a) 21 m (b) 24 m (c) 27 m (d) 18 m. Q2. The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot is 30°. The height of the tower is: (a) 10 m (b) 10√3 m (c) 20 m (d) 20√3 m. Q3. From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. The height of the tower is: (a) 7(√3 + 1) m (b) 7√3 m (c) 14 m (d) 21 m. Q4. A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. The tower is 60 m high. The height of the pole is: (a) 20 m (b) 30 m (c) 40 m (d) 45 m. Q5. If the height of a tower and the distance of the point of observation from its foot are both increased by 10%, then the angle of elevation of its top: (a) increases (b) decreases (c) remains unchanged (d) cannot be determined. Q6. The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. The height of the tower is: (a) 20 m (b) 20√3 m (c) 30 m (d) 40 m.

Section B: Complete each statement with the correct word, phrase or numerical value. Time: 8 minutes. Q7. The angle formed by the line of sight with the horizontal when the object is above the observer's eye is called the angle of __________. Q8. If the angle of elevation of the Sun changes from 30° to 60°, the length of the shadow of a tower decreases by __________ percent (given tan 30° = 1/√3, tan 60° = √3). Q9. From a point on the ground, the angle of elevation of the top of a 10 m tall building is 30°. The distance of the point from the base is __________ m. Q10. The line drawn from the eye of an observer to the point being viewed is called the __________. Q11. If a kite is flying at a height of 60 m from the level ground, attached to a string inclined at 60° to the horizontal, then the length of the string is __________ m. Section C: Match the following or state True/False. Time: 8 minutes. Q12. Angle of depression from point A looking at point B = Angle of elevation from point B looking at point A. (True / False). Q13. Match Column I with Column II: (i) tan 45° → (a) √3, (ii) tan 60° → (b) 1, (iii) tan 30° → (c) 1/√3. Q14. A pole of height h casts a shadow of length h on the ground. The Sun's elevation is 30°. (True / False). Q15. The angle of elevation of the top of a tower increases as we move towards the tower. (True / False). Q16. If the angle of elevation of the Sun is 45°, the height of the tower and length of its shadow are unequal. (True / False). Q17. From the top of a building, the angle of depression of an object on the ground is equal to the angle of elevation of the top of the building from that object. (True / False).

Answer each question in about 40-50 words or 4-5 steps. Show all working. Each question carries 2 marks. Time allocation: 20 minutes. Q18. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 m. Find the height of the tree before it was broken. Q19. The angles of depression of the top and bottom of a 10 m tall building from the top of a tower are 30° and 45° respectively. Find the height of the tower. Q20. From a point on the ground, the angle of elevation of the top of a tower is 30°. On walking 20 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower. Q21. A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 5 m away from the wall, find the length of the ladder. Q22. An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. Find the height of the tower.

Answer in detail with complete steps and diagrams where necessary. Each question carries 5 marks. Time: 30 minutes. Q23. The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Also find the distance between the building and the tower. Q24. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. (Use √3 = 1.732). Q25. From the top of a 100 m high tower, the angles of depression of two cars on the same side of the tower, in the same straight line with the base of the tower, are observed to be 45° and 60°. Find the distance between the two cars. (Give your answer correct to the nearest metre, use √3 = 1.732).

Read the passage carefully and answer the sub-questions that follow. A lighthouse stands vertically on a rocky shore. A ship at sea observes the lighthouse from two positions along a straight line towards the shore. From position A, the angle of elevation of the top of the lighthouse is 30°. After sailing 100 m directly towards the lighthouse, from position B, the angle of elevation becomes 45°. The height of the lighthouse above sea level is h metres, and the distance from position B to the foot of the lighthouse is d metres. Based on this information, answer Q26 to Q29 (1 mark each). Q26. Write the equation relating h and d using the angle of elevation from position B. (a) h = d (b) h = d/√3 (c) h = d√3 (d) h = 2d. Q27. Write the equation relating h and the distance from position A to the foot of the lighthouse. (a) h = (d + 100)/√3 (b) h = (d + 100)√3 (c) h = d + 100 (d) h = (d – 100)/√3. Q28. The value of d (distance from B to foot of lighthouse) is: (a) 50 m (b) 50(√3 + 1) m (c) 50(√3 – 1) m (d) 100 m. Q29. The height h of the lighthouse is: (a) 50 m (b) 50√3 m (c) 50(√3 + 1) m (d) 100 m.

Section A Answers: Q1.(a) 21 m. Use Pythagoras: base on one side = √(15² – 9²) = 12 m; base on other side = √(15² – 12²) = 9 m; width = 12 + 9 = 21 m. Q2.(b) 10√3 m. Height = 30 tan 30° = 30/√3 = 10√3 m. Q3.(a) 7(√3 + 1) m. Let base distance = x. From depression 45°, x = 7. From elevation 60°, top height = 7 + x tan 60° = 7 + 7√3 = 7(1 + √3). Q4.(d) 45 m. Let pole height = h, base distance = d. From depression 30°, d = h√3. From elevation 60°, 60 – h = d tan 60° = d√3 = 3h, so h = 45 m. Q5.(c) remains unchanged. tan θ = h/d = 1.1h/1.1d = h/d, unchanged. Q6.(b) 20√3 m. Let height = h. h/tan 60° + 40 = h/tan 30°, so h/√3 + 40 = h√3, solving gives h = 20√3 m. Section B Answers: Q7. elevation. Q8. The shadow decreases by approximately 66.67%. At 30°: shadow = h√3; at 60°: shadow = h/√3; decrease = h√3 – h/√3 = (3h – h)/√3 = 2h/√3; percentage = (2h/√3)/(h√3) × 100 ≈ 66.67%. Q9. 10√3 m. Distance = 10/tan 30° = 10√3 m. Q10. line of sight. Q11. 40√3 m. Length = 60/sin 60° = 60/(√3/2) = 120/√3 = 40√3 m. Section C Answers: Q12. True. These are alternate angles. Q13. (i)→(b), (ii)→(a), (iii)→(c). Q14. False. If shadow = h, then tan θ = h/h = 1, so θ = 45°, not 30°. Q15. True. As distance decreases, angle increases. Q16. False. At 45°, height = shadow. Q17. True. Alternate angles in parallel horizontal lines. Section D Answers: Q18. Let broken part = x, standing part = y. x cos 30° = 10, so x = 20/√3. y = x sin 30° = 10/√3. Total height = x + y = 30/√3 = 10√3 ≈ 17.32 m. Q19. Let tower height = H, distance from tower base to building base = d. From 45°: H – 10 = d. From 30°: H = d/√3. Solving: H – 10 = H√3, H(√3 – 1) = 10, H ≈ 13.66 m. Q20. Let height = h. h/tan 30° – h/tan 60° = 20, h√3 – h/√3 = 20, h(3 – 1)/√3 = 20, h = 10√3 m. Q21. Length = 5/cos 60° = 5/(1/2) = 10 m. Q22. Height = 1.5 + 28.5 tan 45° = 1.5 + 28.5 = 30 m. Section E Answers: Q23. Let building height = h, distance = d. From tower foot to building top: h = d tan 30° = d/√3. From building foot to tower top: 50 = d tan 60° = d√3. So d = 50/√3 m. Then h = (50/√3)/√3 = 50/3 ≈ 16.67 m. Distance = 28.87 m. Q24. Let pedestal height = h, distance = d. From 45°: h = d. From 60°: h + 1.6 = d√3 = h√3. So h(√3 – 1) = 1.6, h = 1.6/(1.732 – 1) = 1.6/0.732 ≈ 2.19 m. Q25. From 45°: distance of nearer car = 100 m. From 60°: distance of farther car = 100/√3 ≈ 57.74 m. Distance between cars = 100 – 57.74 = 42 m. Case-Study Answers: Q26.(a) h = d. Q27.(a) h = (d + 100)/√3. Q28.(b) 50(√3 + 1) m. From h = d and h = (d + 100)/√3, we get d = (d + 100)/√3, d√3 = d + 100, d(√3 – 1) = 100, d = 100/(√3 – 1) = 50(√3 + 1). Q29.(c) 50(√3 + 1) m, same as d.

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First, keep NCERT examples from Chapter 9 open alongside this worksheet; many question patterns repeat. Draw every diagram even if the question does not ask for one — visual clarity prevents calculation errors. Label all known quantities (heights, distances, angles) directly on the diagram before forming equations. Memorise the six key trigonometric values (sin, cos, tan for 30°, 45°, 60°) on a flashcard; board examiners expect exact answers like 10√3 m, not decimal approximations. For two-angle problems, write two separate equations and solve simultaneously by substitution or elimination. Time yourself: MCQs should take under 1 minute each, short-answer questions 3-4 minutes, long-answer 8-10 minutes. After finishing, compare your method with the answer key — even if your final answer is correct, learn alternate shorter approaches. Retry all incorrect questions after 48 hours without looking at solutions first. This spaced repetition cements concepts. Finally, practice at least three such worksheets before the board exam to achieve speed and accuracy under time pressure.

Mistake 1: Confusing angle of elevation with angle of depression. Remember, elevation is measured upward from horizontal, depression downward from horizontal; both are measured from the observer's eye level, not from the ground. Mistake 2: Using the wrong trigonometric ratio. If you know height and need base, use tan θ = height/base, so base = height/tan θ. Many students mistakenly write base = height × tan θ. Mistake 3: Forgetting to add or subtract observer heights. If a 1.5 m tall person observes a tower, the tower's total height = calculated height + 1.5 m. Mistake 4: Not simplifying surds. Writing 10/√3 instead of 10√3/3 loses marks in board exams; always rationalise denominators. Mistake 5: Misreading 'distance between two objects' as 'distance of one object'. In two-car or two-ship problems, subtract the individual distances from the tower. Mistake 6: Approximating π or √3 too early. Use √3 = 1.732 only at the final step if the question permits decimals; intermediate steps should retain √3. Review these mistakes while checking your worksheet answers and mark any you made with a red pen for focused revision.