Height and Distance Problems
Height and distance problems are practical applications of trigonometry. These problems involve finding the height of a tower, building, cliff, or mountain, or the distance between two objects, using angles of elevation and depression.
In Class 10 CBSE Mathematics (Chapter 9: Some Applications of Trigonometry), these problems require constructing a right-angled triangle from the given situation and applying trigonometric ratios (sin, cos, tan) to find unknown sides or angles.
The two key concepts are:
- Angle of elevation — the angle between the horizontal line of sight and the line of sight directed upward to an object above.
- Angle of depression — the angle between the horizontal line of sight and the line of sight directed downward to an object below.
What is Height and Distance Problems - Trigonometry Applications & Solved Examples?
Definition: Height and distance problems use trigonometric ratios to determine inaccessible heights or distances by measuring angles from a known position.
Key Terms:
- Line of sight — the straight line drawn from the observer's eye to the object
- Horizontal line — the imaginary line from the observer's eye, parallel to the ground
- Angle of elevation (θ) — angle between horizontal and line of sight when looking UP
- Angle of depression (θ) — angle between horizontal and line of sight when looking DOWN
- Observer — the person making the observation
Important:
- The angle of elevation from point A to point B equals the angle of depression from point B to point A (alternate interior angles).
- Unless stated otherwise, assume the observer's height is negligible (observer stands on ground level).
- Standard angles used: 30°, 45°, 60°.
Height and Distance Problems Formula
Trigonometric Ratios Used:
tan θ = Opposite / Adjacent | sin θ = Opposite / Hypotenuse | cos θ = Adjacent / Hypotenuse
Standard Values:
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
Steps to solve height and distance problems:
- Draw a clear diagram with all given information.
- Identify the right triangle(s) in the figure.
- Mark the angle of elevation or depression.
- Choose the appropriate trigonometric ratio (usually tan θ for height/distance problems).
- Substitute known values and solve for the unknown.
Derivation and Proof
Why tan θ is used most often:
- In height and distance problems, the right triangle has a vertical side (height) and a horizontal side (distance).
- tan θ = perpendicular / base = height / distance.
- Since the problem typically involves finding height given distance (or vice versa), tan θ directly relates the two unknowns.
- sin θ and cos θ involve the hypotenuse, which is rarely given or required in these problems.
Angle of depression = Angle of elevation (proof):
- Let AB be a tower. Let C be a point on the ground. Let the observer be at A (top of tower).
- The horizontal at A is parallel to the ground BC.
- The angle of depression at A = angle between horizontal at A and line AC.
- The angle of elevation at C = angle between ground (horizontal at C) and line CA.
- These are alternate interior angles formed by the transversal AC cutting two parallel lines.
- Therefore, angle of depression from A = angle of elevation from C.
Types and Properties
Types of Height and Distance Problems:
- Type 1: Single right triangle — Finding height of a tower/building from a point on the ground (one angle given).
- Type 2: Two positions on the same side — Observer moves closer/farther; two angles of elevation given. Set up two equations with two unknowns.
- Type 3: Angle of depression from top — Observer on top of a building looks down at an object. Use angle of depression = angle of elevation.
- Type 4: Two objects viewed from a point — Finding the distance between two objects seen from the same observation point.
- Type 5: Shadow problems — The sun's angle of elevation determines the length of a shadow. tan θ = height / shadow length.
Solved Examples
Example 1: Finding Height of a Tower
Problem: The angle of elevation of the top of a tower from a point on the ground, 30 m from the foot of the tower, is 60°. Find the height of the tower.
Solution:
Given:
- Distance from foot of tower = 30 m
- Angle of elevation = 60°
Using tan θ:
- tan 60° = height / 30
- √3 = h / 30
- h = 30√3
- h = 30 × 1.732 = 51.96 m
Answer: Height of the tower = 30√3 m ≈ 51.96 m
Example 2: Finding Distance from the Foot
Problem: A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string.
Solution:
Given:
- Height of kite = 75 m
- Angle of elevation = 60°
Using sin θ: (we need hypotenuse = string length)
- sin 60° = 75 / string length
- √3/2 = 75 / L
- L = 75 × 2/√3 = 150/√3
- L = 150√3/3 = 50√3
- L = 50 × 1.732 = 86.6 m
Answer: Length of the string = 50√3 m ≈ 86.6 m
Example 3: Angle of Depression from a Building
Problem: From the top of a 45 m high building, the angle of depression of a car on the ground is 30°. Find the distance of the car from the base of the building.
Solution:
Given:
- Height of building = 45 m
- Angle of depression = 30°
Note: Angle of depression = Angle of elevation from the car = 30°
Using tan θ:
- tan 30° = 45 / d
- 1/√3 = 45 / d
- d = 45√3
- d = 45 × 1.732 = 77.94 m
Answer: Distance of car from base = 45√3 m ≈ 77.94 m
Example 4: Shadow Length Problem
Problem: The shadow of a tower standing on level ground is found to be 40 m longer when the sun's altitude (angle of elevation) is 30° than when it is 60°. Find the height of the tower.
Solution:
Let: Height of tower = h, Shadow at 60° = x m
At 60° elevation:
- tan 60° = h/x → √3 = h/x → x = h/√3 ... (1)
At 30° elevation: shadow = x + 40
- tan 30° = h/(x + 40) → 1/√3 = h/(x + 40) → x + 40 = h√3 ... (2)
Substituting (1) in (2):
- h/√3 + 40 = h√3
- 40 = h√3 − h/√3
- 40 = h(√3 − 1/√3) = h(3 − 1)/√3 = 2h/√3
- h = 40√3/2 = 20√3
- h = 20 × 1.732 = 34.64 m
Answer: Height of tower = 20√3 m ≈ 34.64 m
Example 5: Two Angles of Elevation from Different Points
Problem: The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower, in the same straight line, are complementary. Find the height of the tower.
Solution:
Let: Height = h, angles = θ and (90° − θ)
From point at 9 m:
- tan θ = h/9 ... (1)
From point at 4 m:
- tan(90° − θ) = h/4 → cot θ = h/4 → 1/tan θ = h/4 ... (2)
Multiplying (1) and (2):
- tan θ × (1/tan θ) = (h/9) × (h/4)
- 1 = h²/36
- h² = 36
- h = 6 m
Answer: Height of tower = 6 m
Example 6: Distance Between Two Ships
Problem: From the top of a 100 m high lighthouse, the angles of depression of two ships on opposite sides are 30° and 45°. Find the distance between the two ships.
Solution:
Given: Height of lighthouse = 100 m
For Ship 1 (angle of depression 30°):
- tan 30° = 100/d₁
- 1/√3 = 100/d₁
- d₁ = 100√3 m
For Ship 2 (angle of depression 45°):
- tan 45° = 100/d₂
- 1 = 100/d₂
- d₂ = 100 m
Distance between ships:
- = d₁ + d₂ (ships on opposite sides)
- = 100√3 + 100
- = 100(√3 + 1)
- = 100(1.732 + 1) = 273.2 m
Answer: Distance between ships = 100(√3 + 1) m ≈ 273.2 m
Example 7: Height of a Cliff from a Boat
Problem: A man in a boat rowing away from a cliff 150 m high takes 2 minutes to change the angle of elevation of the top of the cliff from 60° to 45°. Find the speed of the boat.
Solution:
At angle 60°:
- tan 60° = 150/x → x = 150/√3 = 50√3 m
At angle 45°:
- tan 45° = 150/y → y = 150 m
Distance travelled: y − x = 150 − 50√3 = 150 − 86.6 = 63.4 m
Speed: Distance / Time = 63.4 / 2 = 31.7 m/min
Answer: Speed of boat = (150 − 50√3)/2 m/min ≈ 31.7 m/min
Example 8: Height of a Pole on a Building
Problem: A flagstaff stands on the top of a 5 m high tower. From a point on the ground, the angle of elevation of the top of the flagstaff is 60° and the angle of elevation of the top of the tower is 45°. Find the height of the flagstaff.
Solution:
Let: Height of flagstaff = h, distance from base = d
For tower top (45°):
- tan 45° = 5/d → d = 5 m
For flagstaff top (60°):
- tan 60° = (5 + h)/5
- √3 = (5 + h)/5
- 5√3 = 5 + h
- h = 5√3 − 5 = 5(√3 − 1)
- h = 5(1.732 − 1) = 5 × 0.732 = 3.66 m
Answer: Height of flagstaff = 5(√3 − 1) m ≈ 3.66 m
Example 9: Two Buildings Facing Each Other
Problem: Two buildings are facing each other on either side of a road of width 10 m. From the top of the first building, which is 30 m high, the angle of elevation of the top of the second building is 45°. Find the height of the second building.
Solution:
Given: Building 1 height = 30 m, Road width = 10 m, Angle of elevation = 45°
Let: Height of Building 2 = H
The elevation of 45° is from top of Building 1 to the top of Building 2:
- Vertical difference = H − 30
- Horizontal distance = 10 m
Using tan 45°:
- tan 45° = (H − 30)/10
- 1 = (H − 30)/10
- H − 30 = 10
- H = 40 m
Answer: Height of second building = 40 m
Example 10: Person Moving Toward a Tower
Problem: A person walking towards a tower observes that the angle of elevation of the top changes from 30° to 60° after walking 20 m. Find the height of the tower and the remaining distance to the tower.
Solution:
Let: Height = h, remaining distance after walking = x
At 60° (closer point):
- tan 60° = h/x → √3 = h/x → x = h/√3 ... (1)
At 30° (farther point, distance = x + 20):
- tan 30° = h/(x + 20) → 1/√3 = h/(x + 20) → x + 20 = h√3 ... (2)
Substituting (1) in (2):
- h/√3 + 20 = h√3
- 20 = h√3 − h/√3 = h(3 − 1)/√3 = 2h/√3
- h = 20√3/2 = 10√3 = 17.32 m
Remaining distance: x = h/√3 = 10√3/√3 = 10 m
Answer: Height = 10√3 m ≈ 17.32 m, Remaining distance = 10 m
Real-World Applications
Real-life applications of height and distance:
- Surveying: Engineers use theodolites to measure angles of elevation and calculate heights of mountains, towers, and buildings.
- Navigation: Pilots and sailors determine distances and altitudes using trigonometric calculations.
- Architecture: Architects calculate building heights, shadow lengths, and optimal placement using trigonometry.
- Astronomy: Distances to stars and heights of celestial objects are estimated using angular measurements.
- Military: Range finding and targeting use height and distance calculations.
- Construction: Crane operators and construction workers calculate reach distances and elevation angles for safe operations.
Key Points to Remember
- Angle of elevation is measured upward from the horizontal; angle of depression is measured downward.
- Angle of depression from A to B = angle of elevation from B to A (alternate interior angles).
- tan θ is the most frequently used ratio: tan θ = height / distance.
- Always draw a clear diagram before solving.
- Standard angles: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
- Shadow problems: tan(sun's altitude) = height of object / length of shadow.
- When two angles are complementary (sum = 90°), use the identity tan θ × tan(90° − θ) = 1.
- For problems with two triangles, set up two equations and eliminate the common unknown.
- Heights are always positive — reject negative solutions.
- Use √3 = 1.732 for final numerical answers unless the problem asks for exact form.
Practice Problems
- From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top is 45°. Find the height of the tower.
- The angle of elevation of the top of a tree from a point at a distance of 200 m is 30°. Find the height of the tree.
- A 1.5 m tall boy stands at some distance from a 30 m high building. The angle of elevation from his eyes to the top of the building is 60°. Find the distance from the boy to the building.
- 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°. Find the height of the tower.
- The shadow of a vertical pole is 1/√3 times its height. Find the angle of elevation of the sun.
- Two poles of equal heights are standing opposite each other 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 height of the poles and the position of the point.
Frequently Asked Questions
Q1. What are height and distance problems?
Height and distance problems use trigonometric ratios to find the height of objects (towers, buildings, mountains) or distances between objects, given the angle of elevation or depression from a known point.
Q2. Which trigonometric ratio is used most in height and distance problems?
tan θ is used most often because it relates the perpendicular (height) to the base (distance), which are the two quantities typically involved: tan θ = height / distance.
Q3. What is the difference between angle of elevation and angle of depression?
Angle of elevation is the angle formed when looking UP from the horizontal line. Angle of depression is formed when looking DOWN from the horizontal line. If A is above B, the angle of depression from A = angle of elevation from B.
Q4. Can the angle of elevation be 90°?
Practically, no. An angle of elevation of 90° means the object is directly overhead, which makes the horizontal distance zero. This is a degenerate case not used in standard problems.
Q5. How do you solve problems with two angles of elevation?
Set up two equations using tan θ for each angle. Both equations share the height (h) as a common variable. Solve simultaneously by substituting one into the other to find h and the distance.
Q6. What values of √3 and 1/√3 should I use?
√3 ≈ 1.732 and 1/√3 ≈ 0.577. In board exams, leave answers in surd form (e.g., 20√3 m) unless the problem specifically asks for a decimal approximation.
Q7. What if the observer has a height?
If the observer's height is given (e.g., a 1.5 m tall boy), subtract the observer's height from the object's height to get the effective vertical distance used in the trigonometric ratio.
Q8. How do complementary angles help in these problems?
If two angles of elevation from two points are complementary (α + β = 90°), then tan α × tan β = 1. This simplifies the calculation: h² = d₁ × d₂, so h = √(d₁ × d₂).
