NCERT Class 10 Mathematics | Chapter 9 | CBSE Board (2020–2025)
This chapter applies trigonometric ratios to solve real-world problems involving heights and distances. Key concepts include angle of elevation, angle of depression, and the use of right-angled triangles to find unknown heights or distances without direct measurement.
Exam Relevance: Consistently 4–6 marks (1 MCQ + 1 short + 1 long answer). Frequently tested in both Main and Compartment Exams (2020–2025).
Sourced from Main & Compartment Exams, Sample Papers, and NCERT Exemplar
Q1. If the angle of elevation of the top of a tower from a point 30 m away from its foot is 30°, then the height of the tower is:
Answer: B
Q2. The angle of depression of a boat from the top of a 50 m high lighthouse is 45°. The distance of the boat from the lighthouse is:
Answer: B
Q3. A kite is flying at a height of 60 m above the ground. The string attached to the kite is inclined at 60° to the ground. The length of the string is:
Answer: A
Q4. If the sun’s altitude is 60°, the length of the shadow of a 10 m high pole is:
Answer: B
Q5. The angle of elevation of the top of a tower from two points at distances a and b from the base (on the same side) are complementary. The height of the tower is:
Answer: B
Q6. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. The height of the tower is:
Answer: A
Q7. A ladder 10 m long reaches a window 8 m above the ground. The distance of the foot of the ladder from the wall is:
Answer: B
Q8. The angle of elevation of the sun when the shadow of a pole is √3 times its height is:
Answer: A
Q9. A tree breaks due to storm and the broken part bends so that the top 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 is 8 m. The height of the tree was:
Answer: A
Q10. The upper part of a tree broken by wind makes an angle of 60° with the ground. The distance from the root to the point where the top touches is 4 m. The original height of the tree is:
Answer: D
Q11. If the angle of elevation of a tower is 45° at a distance of 20 m, its height is:
Answer: B
Q12. A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he moves 40 m away, the angle becomes 30°. The width of the river is:
Answer: A
Q13. The length of the shadow of a vertical pole when the sun’s altitude is 30° is 15 m. The height of the pole is:
Answer: A
Q14. From a point on the ground 20 m away from the foot of a tower, the angle of elevation is 45°. The height of the tower is:
Answer: B
Q15. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top is 30°. The distance is:
Answer: A
Q16. The angle of depression of a car from the top of a 50 m high tower is 30°. The distance of the car from the tower is:
Answer: B
Q17. A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 2.5 m from the wall, its length is:
Answer: A
Q18. If the height of a vertical pole is equal to the length of its shadow, the angle of elevation of the sun is:
Answer: B
Q19. The angle of elevation of the top of a tower from a point 60 m away is 30°. The height of the tower is:
Answer: A
Q20. A person standing 30 m from a tower finds the angle of elevation to be 60°. The height of the tower is:
Answer: B
Q21. Define angle of elevation. (1 mark)
The angle formed by the line of sight with the horizontal when the object is above the horizontal level is called the angle of elevation.
Marks: 1 for correct definition
Q22. The angle of elevation of the top of a tower from a point 20 m away is 45°. Find its height. (2 marks)
Let height = h.
tan 45° = h / 20
⇒ 1 = h / 20
⇒ h = 20 m
Marks: 1 for equation, 1 for answer
Q23. A ladder 10 m long reaches a window 6 m above the ground. Find the distance of its foot from the wall. (2 marks)
Let distance = x.
By Pythagoras: x = √(10² – 6²) = √(100 – 36) = √64 = 8 m
Marks: 1 for formula, 1 for answer
Q24. The length of the shadow of a 6 m high pole is 2√3 m. Find the angle of elevation of the sun. (2 marks)
tan θ = height / shadow = 6 / (2√3) = 3/√3 = √3
⇒ θ = 60°
Marks: 1 for tan θ, 1 for angle
Q25. From a point on the ground, the angle of elevation of the top of a tower is 30°. If the tower is 50 m high, find the distance of the point from the tower. (2 marks)
tan 30° = 50 / d ⇒ 1/√3 = 50/d ⇒ d = 50√3 m
Marks: 1 for equation, 1 for answer
Q26. The angle of depression of a boat from a 45 m high cliff is 60°. Find the distance of the boat from the cliff. (2 marks)
Angle of elevation = 60°.
tan 60° = 45 / d ⇒ √3 = 45/d ⇒ d = 45/√3 = 15√3 m
Marks: 1 for relation, 1 for answer
Q27. A 1.2 m tall girl spots a balloon at a height of 88.2 m. The angle of elevation is 60°. Find the distance from the girl to the balloon. (3 marks)
Effective height = 88.2 – 1.2 = 87 m
tan 60° = 87 / d ⇒ √3 = 87/d ⇒ d = 87/√3 = 29√3 m
Marks: 1 for effective height, 1 for equation, 1 for answer
Q28. The angles of elevation of the top of a tower from two points at distances 4 m and 9 m are complementary. Find the height of the tower. (3 marks)
Let angles be θ and (90° – θ).
h = 4 tan θ and h = 9 cot θ
⇒ h² = 36 ⇒ h = 6 m
Marks: 1 for equations, 1 for multiplication, 1 for answer
Q29. A tree breaks and the top touches the ground at 30°, 10 m from the foot. Find original height. (3 marks)
Let broken part = l, standing part = h.
cos 30° = 10/l ⇒ l = 20/√3
tan 30° = h/10 ⇒ h = 10/√3
Total height = h + l = 30/√3 = 10√3 m
Marks: 1 for l, 1 for h, 1 for total
Q30. From a 20 m high building, angle of elevation to a tower top is 30° and depression to its foot is 45°. Find tower height. (3 marks)
Distance = 20 m (from depression 45°).
Extra height = 20 tan 30° = 20/√3 m.
Total height = 20 + 20/√3 = 20(1 + 1/√3) m
Marks: 1 for distance, 1 for extra height, 1 for total
Q31. A vertical pole of height 12 m casts a shadow 4√3 m long. Find the angle of elevation of the sun. (2 marks)
tan θ = 12 / (4√3) = 3/√3 = √3 ⇒ θ = 60°
Marks: 1 for ratio, 1 for angle
Q32. From a point 30 m away from a tower, the angle of elevation is 60°. Find the height. (2 marks)
tan 60° = h/30 ⇒ h = 30√3 m
Marks: 1 for equation, 1 for answer
Q33. The angle of depression of a car from a 25 m high building is 30°. Find distance of car. (2 marks)
tan 30° = 25/d ⇒ d = 25√3 m
Marks: 1 for equation, 1 for answer
Q34. A 1.7 m tall man is 50 m away from a tower. Angle of elevation to top is 45°. Find tower height. (3 marks)
Effective height = h – 1.7
tan 45° = (h – 1.7)/50 ⇒ h – 1.7 = 50 ⇒ h = 51.7 m
Marks: 1 for effective height, 1 for equation, 1 for total
Q35. Two poles of height 6 m and 11 m stand on a plane ground. If distance between feet is 12 m, find distance between tops. (3 marks)
Height diff = 5 m, base = 12 m.
Distance = √(12² + 5²) = √169 = 13 m
Marks: 1 for diff, 1 for Pythagoras, 1 for answer
Q36. A man observes a tower at 30°. After walking 40 m towards it, angle becomes 60°. Find height. (3 marks)
Let final distance = x.
h = x tan 60° = x√3; also h = (x+40) tan 30° = (x+40)/√3
⇒ x√3 = (x+40)/√3 ⇒ 3x = x + 40 ⇒ x = 20 ⇒ h = 20√3 m
Marks: 1 for equations, 1 for solving, 1 for h
Q37. The shadow of a tower is 50 m when sun’s elevation is 30°. What is height? (2 marks)
tan 30° = h/50 ⇒ h = 50/√3 m
Marks: 1 for equation, 1 for answer
Q38. From a 10 m high cliff, angles of depression of two boats are 45° and 30°. Find distance between boats if on same side. (3 marks)
d₁ = 10 m (45°), d₂ = 10√3 m (30°)
Distance = d₂ – d₁ = 10(√3 – 1) m
Marks: 1 for each distance, 1 for difference
Q39. A balloon is connected to a station by a 200 m string inclined at 60°. Find height. (2 marks)
sin 60° = h/200 ⇒ h = 200 × √3/2 = 100√3 m
Marks: 1 for ratio, 1 for answer
Q40. If angle of elevation of cloud from a point 60 m above lake is 30° and angle of depression of reflection is 60°, find cloud height. (3 marks)
Let cloud height above lake = h.
Then: (h – 60)/d = tan 30°, (h + 60)/d = tan 60°
Divide: (h + 60)/(h – 60) = 3 ⇒ h + 60 = 3h – 180 ⇒ 2h = 240 ⇒ h = 120 m
Marks: 1 for equations, 1 for division, 1 for answer
Q41. From the top of a 60 m high lighthouse, the angles of depression of two ships are 30° and 45°. If the ships are on the same side, find the distance between them. (4 marks)
Let distances be d₁ and d₂ (d₁ > d₂).
tan 45° = 60/d₂ ⇒ d₂ = 60 m
tan 30° = 60/d₁ ⇒ d₁ = 60√3 m
Distance between ships = d₁ – d₂ = 60√3 – 60 = 60(√3 – 1) m
Marks: 1 for d₂, 1 for d₁, 1 for subtraction, 1 for final answer
Common Error: Using sum instead of difference.
Q42. The angle of elevation of the top of a tower from a point A is 30°. On moving 50 m towards the tower to point B, the angle becomes 60°. Find the height of the tower and its distance from A. (5 marks)
Let height = h, distance from B = x.
From B: tan 60° = h/x ⇒ h = x√3 …(1)
From A: tan 30° = h/(x + 50) ⇒ h = (x + 50)/√3 …(2)
Equate (1) & (2): x√3 = (x + 50)/√3 ⇒ 3x = x + 50 ⇒ x = 25 m
Then h = 25√3 m
Distance from A = x + 50 = 75 m
Marks: 1 for each equation, 1 for solving, 1 for h, 1 for distance
Q43. A vertical tower is surmounted by a flagstaff of height 7 m. From a point on the ground, the angles of elevation of the bottom and top of the flagstaff are 30° and 45° respectively. Find the height of the tower. (4 marks)
Let tower height = h, distance = d.
tan 30° = h/d ⇒ d = h√3 …(1)
tan 45° = (h + 7)/d ⇒ d = h + 7 …(2)
From (1) & (2): h√3 = h + 7 ⇒ h(√3 – 1) = 7 ⇒ h = 7/(√3 – 1) = (7(√3 + 1))/2 m
Marks: 1 for each equation, 1 for solving, 1 for rationalized answer
Q44. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower on a 20 m high building are 45° and 60°. Find tower height. (4 marks)
Distance = 20 m (from 45°).
Total height = 20 tan 60° = 20√3 m
Tower height = 20√3 – 20 = 20(√3 – 1) m
Marks: 1 for distance, 1 for total height, 1 for subtraction, 1 for answer
Q45. A 1.5 m tall boy is standing at a distance from a 30 m tall building. He walks towards it. When angle of elevation changes from 30° to 60°, find distance walked. (5 marks)
Effective height = 28.5 m
Initial distance: d₁ = 28.5 / tan 30° = 28.5√3 m
Final distance: d₂ = 28.5 / tan 60° = 28.5/√3 m
Distance walked = d₁ – d₂ = 28.5(√3 – 1/√3) = 28.5(2/√3) = 19√3 m
Marks: 1 for effective height, 1 each for d₁ & d₂, 1 for subtraction, 1 for simplification
Q46. The angle of elevation of a cloud from a point 120 m above a lake is 30° and the angle of depression of its reflection is 60°. Find cloud height. (5 marks)
Let cloud height above lake = h.
Then: (h – 120)/d = tan 30° = 1/√3 …(1)
(h + 120)/d = tan 60° = √3 …(2)
Divide (2) by (1): (h + 120)/(h – 120) = 3
⇒ h + 120 = 3h – 360 ⇒ 2h = 480 ⇒ h = 240 m
Marks: 1 for each equation, 1 for division, 1 for solving, 1 for answer
Q47. From the top of a 75 m high tower, the angles of depression of two cars are 30° and 45° on opposite sides. Find distance between cars. (4 marks)
d₁ = 75 / tan 30° = 75√3 m
d₂ = 75 / tan 45° = 75 m
Total distance = d₁ + d₂ = 75√3 + 75 = 75(√3 + 1) m
Marks: 1 for each distance, 1 for sum, 1 for answer
Q48. A tree breaks and the top touches ground at 60°, 8 m from root. Find original height. (4 marks)
Let standing part = h, broken part = l.
tan 60° = h/8 ⇒ h = 8√3
cos 60° = 8/l ⇒ l = 16
Total height = 8√3 + 16 = 8(2 + √3) m
Marks: 1 for h, 1 for l, 1 for total, 1 for factorisation
Q49. From a point on a bridge 30 m above river, angles of depression of banks are 30° and 45°. Find river width. (5 marks)
Width = d₁ + d₂
d₁ = 30 / tan 30° = 30√3 m
d₂ = 30 / tan 45° = 30 m
Total = 30√3 + 30 = 30(√3 + 1) m
Marks: 1 for diagram idea, 1 for each distance, 1 for sum, 1 for answer
Q50. A ladder rests against a wall at 60°. Its foot is pulled away 2 m, and it slips 2 m. Find original length. (5 marks)
Let original foot distance = x, height = y, length = L.
Initially: x = L cos 60° = L/2, y = L sin 60° = L√3/2
After: (x + 2)² + (y – 2)² = L²
Substitute: (L/2 + 2)² + (L√3/2 – 2)² = L²
Expand: L²/4 + 2L + 4 + 3L²/4 – 2L√3 + 4 = L²
Simplify: L² + 8 – 2L√3 = L² ⇒ 8 = 2L√3 ⇒ L = 4/√3 = (4√3)/3 m
Marks: 1 for initial, 1 for new positions, 1 for equation, 1 for expansion, 1 for answer