Chapter 9: Some Applications of Trigonometry
NCERT Solutions for CBSE Class 10 Mathematics — 20 solved questions with detailed explanations.
Important Formulas
Solved Questions
Q1. The angle of depression and angle of elevation between two points are:
Difficulty: Easy · Topic: Heights and Distances
Solution
By alternate interior angles (horizontal lines are parallel), the angle of depression from A to B equals the angle of elevation from B to A.
Q2. The angle of elevation of the top of a tower from a point 100 m away is 30°. Find the height of the tower.
Difficulty: Easy-Medium · Topic: Angle of Elevation
Solution
tan 30° = h/100 → h = 100 × 1/√3 = 100√3/3 ≈ 57.74 m.
Q3. From the top of a 75 m high building, the angle of depression of a car is 30°. Find the distance of the car from the base.
Difficulty: Easy-Medium · Topic: Angle of Depression
Solution
tan 30° = 75/d → d = 75/tan30° = 75√3 m ≈ 129.9 m.
Q4. A tower is 100 m high. Angle of elevation from a point is 45°. Distance from base is:
Difficulty: Easy-Medium · Topic: Angle of Elevation
Solution
tan45° = 100/d → 1 = 100/d → d = 100 m.
Q5. A tree breaks due to storm and the broken part bends so that the top touches the ground making 30° with it. If the point is 8 m from the foot, find the height of the tree.
Difficulty: Easy-Medium · Topic: Angle of Elevation
Solution
Standing part = 8 tan30° = 8/√3. Broken part = 8/cos30° = 16/√3.
Total = 8/√3 + 16/√3 = 24/√3 = 8√3 m.
Q6. The angle of elevation of the top of a tower from two points at distances a and b (a > b) from the base are 30° and 60° respectively. Find the height.
Difficulty: Medium · Topic: Angle of Elevation
Solution
h = a tan30° = a/√3. Also h = b tan60° = b√3.
a/√3 = b√3 → a = 3b. h = a/√3 = 3b/√3 = b√3.
h² = (a/√3)(b√3) = ab. So h = √(ab).
Q7. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and angle of depression of its foot is 45°. Find the height of the tower.
Difficulty: Medium · Topic: Heights and Distances
Solution
Let distance = d. tan45° = 7/d → d = 7 m.
Let extra height = h. tan60° = h/7 → h = 7√3.
Total tower height = 7 + 7√3 = 7(1+√3) m.
Q8. A kite is flying at height 60 m. The string makes 60° with the horizontal. Find the length of the string (assume no sag).
Difficulty: Medium · Topic: Angle of Elevation
Solution
sin 60° = 60/L → √3/2 = 60/L → L = 120/√3 = 40√3 m.
Q9. A pole 6 m high casts a shadow 2√3 m long. Find the angle of elevation of the sun.
Difficulty: Medium · Topic: Heights and Distances
Solution
tan θ = 6/(2√3) = 3/√3 = √3 → θ = 60°.
Q10. Two poles of equal heights stand on either side of a road 80 m wide. From a point between them on the road, the angles of elevation of the tops are 60° and 30°. Find the heights and the position of the point.
Difficulty: Medium · Topic: Angle of Elevation
Solution
Let height = h, point at distance d from first pole.
tan60° = h/d → h = d√3. tan30° = h/(80−d) → h = (80−d)/√3.
d√3 = (80−d)/√3 → 3d = 80−d → 4d = 80 → d = 20.
h = 20√3 m.
Q11. A 1.5 m tall observer is 28.5 m from a chimney. The angle of elevation of the chimney top from her eyes is 45°. Find the chimney's height.
Difficulty: Medium · Topic: Angle of Depression
Solution
Extra height above eye = 28.5 × tan45° = 28.5 m. Total = 28.5 + 1.5 = 30 m.
Q12. An aeroplane at height 300 m sees angles of depression of two points on opposite banks of a river as 45° and 60°. Find the width of the river.
Difficulty: Medium · Topic: Heights and Distances
Solution
Width = 300/tan45° + 300/tan60° = 300 + 300/√3 = 300(1+1/√3) ≈ 473.2 m.
Q13. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower on a building are 45° and 60°. If the building is 15 m high, find the tower height.
Difficulty: Medium · Topic: Heights and Distances
Solution
tan45° = 15/d → d = 15. Let tower = h. tan60° = (15+h)/15 → √3 = 1+h/15 → h = 15(√3−1).
Q14. The shadow of a tower on level ground is found to be 40 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.
Difficulty: Medium · Topic: Angle of Depression
Solution
h/tan30° − h/tan60° = 40. h√3 − h/√3 = 40 → h(3−1)/√3 = 40 → 2h/√3 = 40 → h = 20√3.
Q15. 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 of the top of the pedestal is 45°. Find the height of the pedestal.
Difficulty: Medium · Topic: Angle of Elevation
Solution
Let pedestal = h, distance = d. tan45° = h/d → d = h. tan60° = (h+1.6)/h → √3 = 1+1.6/h → h = 1.6/(√3−1) = 0.8(√3+1).
Q16. From the top of a 60 m high tower, the angles of depression of the top and bottom of a building are 30° and 60° respectively. Find the height of the building.
Difficulty: Medium · Topic: Heights and Distances
Solution
tan60° = 60/d → d = 60/√3 = 20√3. tan30° = (60−h)/d → 1/√3 = (60−h)/(20√3) → 20 = 60−h → h = 40 m.
Q17. The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are 30° and 45°. Find the height of the tower and distance between them.
Difficulty: Medium-Hard · Topic: Heights and Distances
Solution
Let tower height = h, distance = d.
tan45° = h/d → d = h.
tan30° = (h−8)/d → 1/√3 = (h−8)/h → h = √3(h−8).
h = √3h−8√3 → h(√3−1) = 8√3 → h = 8√3/(√3−1) = 8√3(√3+1)/2 = 4√3(√3+1) = 4(3+√3) = 12+4√3.
Hmm, let me redo: h = 8√3/(√3−1) × (√3+1)/(√3+1) = 8√3(√3+1)/2 = 4√3(√3+1) = 4(3+√3) ≈ 18.93.
d = h ≈ 18.93 m.
Q18. A straight highway leads to the foot of a tower. A man standing at the top observes a car at angle of depression 30°, which is approaching at uniform speed. Six seconds later the angle of depression is 60°. Find the time taken by the car to reach the foot of the tower.
Difficulty: Medium-Hard · Topic: Angle of Elevation
Solution
Let h = tower height. At 30°: d₁ = h√3. At 60°: d₂ = h/√3.
Distance covered in 6 s = d₁−d₂ = h√3−h/√3 = 2h/√3.
Speed = 2h/(√3×6) = h/(3√3).
Remaining distance = h/√3. Time = (h/√3)/(h/(3√3)) = 3 s.
Q19. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite, the angle of elevation of the top is 60°. From another point 20 m away from this point on the same bank, the angle of elevation is 30°. Find the height of the tower and width of the canal.
Difficulty: Medium-Hard · Topic: Heights and Distances
Solution
tan60° = h/w → h = w√3. tan30° = h/(w+20) → h = (w+20)/√3.
w√3 = (w+20)/√3 → 3w = w+20 → 2w = 20 → w = 10. h = 10√3.
Q20. A vertical tower stands on a horizontal plane. From a point on the ground the angle of elevation of the top is tan⁻¹(5/12). On walking 192 m towards the tower, it becomes tan⁻¹(3/4). Find the height.
Difficulty: Hard · Topic: Heights and Distances
Solution
Let h = height, d = initial distance. h/d = 5/12 → d = 12h/5. h/(d−192) = 3/4 → d−192 = 4h/3.
12h/5 − 192 = 4h/3 → (36h−20h)/15 = 192 → 16h = 2880 → h = 180 m.
Other Chapters in Mathematics
Want to practice these questions interactively?
Get instant feedback, track progress, and improve with adaptive practice.
Start Practicing Free →
