Heights and Distances
Introduction
Heights and distances is the practical application of trigonometry to measure inaccessible heights and distances. In ICSE Class 10, you use trigonometric ratios to solve problems involving angles of elevation and depression.
Key Terms
- Angle of Elevation — The angle between the horizontal line of sight and the line joining the observer's eye to an object ABOVE the observer.
- Angle of Depression — The angle between the horizontal line of sight and the line joining the observer's eye to an object BELOW the observer.
Basic Approach
- Draw a well-labelled diagram representing the problem.
- Identify the right triangle(s) in the diagram.
- Mark the known angles and sides.
- Choose the appropriate trigonometric ratio (usually tan or sin).
- Set up the equation and solve for the unknown.
Standard Angle Values
| Angle | tan | sin | cos |
|---|---|---|---|
| 30° | ¹/√3 | ¹/₂ | √3/2 |
| 45° | 1 | ¹/√2 | ¹/√2 |
| 60° | √3 | √3/2 | ¹/₂ |
Worked Examples
Example 1: Angle of Elevation
A tower casts a shadow 50 m long when the sun's altitude is 30°. Find the height of the tower.
Solution: Let height of tower = h m. In the right triangle: tan 30° = h / 50 ¹/√3 = h / 50 h = 50 / √3 = 50√3 / 3 ≈ 28.87 m
Example 2: Two Angles of Elevation
From a point on the ground, the angle of elevation of the top of a building is 60°. On moving 20 m closer, the angle becomes 75°. Find the height of the building. (Given tan 75° = 2 + √3)
Solution: Let height = h, initial distance from building = x. First position: tan 60° = h / x → √3 = h / x → x = h / √3 Second position: tan 75° = h / (x − 20) → 2 + √3 = h / (h/√3 − 20)
2 + √3 = h / (h/√3 − 20) (2 + √3)(h/√3 − 20) = h (2 + √3)h/√3 − 20(2 + √3) = h h[(2 + √3)/√3 − 1] = 20(2 + √3) h[(2 + √3 − √3)/√3] = 20(2 + √3) h[2/√3] = 20(2 + √3) h = 20√3(2 + √3) / 2 h = 10√3(2 + √3) = 20√3 + 30
Height ≈ 20 × 1.732 + 30 = 34.64 + 30 = 64.64 m
Example 3: Angle of Depression
From the top of a lighthouse 75 m high, the angle of depression of a ship is 30°. Find the distance of the ship from the lighthouse.
Solution: Angle of depression = angle of elevation (alternate angles). Let distance = x m. tan 30° = 75 / x ¹/√3 = 75 / x x = 75√3 ≈ 129.9 m
Example 4: Two Objects on Opposite Sides
From the top of a 60 m high tower, the angles of depression of two cars on the ground on opposite sides of the tower are 30° and 60°. Find the distance between the cars.
Solution: Let the cars be at distances x₁ and x₂ from the tower.
For car 1: tan 30° = 60 / x₁ ¹/√3 = 60 / x₁ x₁ = 60√3 m
For car 2: tan 60° = 60 / x₂ √3 = 60 / x₂ x₂ = 60 / √3 = 20√3 m
Distance between cars = x₁ + x₂ = 60√3 + 20√3 = 80√3 m ≈ 138.56 m
Example 5: Height of Cloud
The angle of elevation of a cloud from a point 50 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud above the lake.
Solution: Let height of cloud above the lake = h m. Observer is at 50 m above the lake.
In right triangle with cloud: tan 30° = (h − 50) / x ¹/√3 = (h − 50) / x x = √3(h − 50) ...(i)
In right triangle with reflection: tan 60° = (h + 50) / x √3 = (h + 50) / x ...(ii)
From (i) and (ii): √3(h − 50) = (h + 50) / √3 3(h − 50) = h + 50 3h − 150 = h + 50 2h = 200 h = 100 m
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Confusing elevation and depression | Elevation: object above; Depression: object below |
| Not drawing a diagram | Always draw before solving |
| Using the wrong trigonometric ratio | Choose tan when opposite and adjacent are involved |
| Angle of depression not equal to angle of elevation | They are equal by alternate angle property (when measured from same horizontal) |
ICSE Exam Focus
Heights and distances carry 6–8 marks in ICSE exams (usually one long question). Types:
- Single observation (find height or distance).
- Two observations with movement.
- Two objects on same/opposite sides.
- Cloud/reflection problems (advanced).
Marks Blueprint:
| Topic | Marks |
|---|---|
| Single angle of elevation/depression | 3 |
| Two-angle problems (movement) | 4 |
| Two objects (opposite/same side) | 3 |
| Advanced problems (reflection, flags) | 4 |
Self-Test Questions
-
A vertical pole 10 m high casts a shadow 10√3 m long. Find the angle of elevation of the sun.
-
The angle of elevation of the top of a tower from a point 100 m away is 45°. Find the height of the tower.
-
From the top of a cliff 80 m high, the angles of depression of two boats at sea are 30° and 60°. If both boats are on the same side of the cliff, find the distance between them.
-
A man 2 m tall observes that the angle of elevation of the top of a pole 20 m away is 60°. Find the height of the pole.
-
The angle of elevation of a jet plane from a point on the ground changes from 30° to 60° in 15 seconds. If the plane is flying at a height of 1500√3 m, find its speed.
In ICSE, draw a clear labelled diagram and show all trigonometric steps. Marks are given for correct angle identification even if the final arithmetic has a minor error.
