Applications of Trigonometry
Applications of Trigonometry deal with using trigonometric ratios to find heights and distances that cannot be measured directly. This topic is covered in Chapter 9 (Some Applications of Trigonometry) of the NCERT Class 10 textbook.
The two key concepts are:
- Angle of elevation: The angle above the horizontal when looking up at an object.
- Angle of depression: The angle below the horizontal when looking down at an object.
Using these angles along with trigonometric ratios (sin, cos, tan), we can calculate heights of buildings, towers, mountains, widths of rivers, and distances between objects.
This chapter represents the practical side of trigonometry. While Chapter 8 builds the theoretical foundation (ratios, identities), Chapter 9 shows how these ratios are applied to solve real-world measurement problems. Historically, trigonometry was developed precisely for this purpose — ancient astronomers and surveyors needed to measure distances and heights that were physically inaccessible.
Problems in this chapter follow a consistent pattern: identify the right triangle in the given physical situation, set up the appropriate trigonometric equation using the given angle of elevation or depression, and solve for the unknown quantity. Most problems use the tangent ratio because the typical setup involves a vertical height and a horizontal distance, which are the opposite and adjacent sides respectively.
CBSE board exams typically include one or two problems from this chapter, carrying 4-5 marks each. These problems require a clear diagram, correct identification of angles, and accurate computation. The standard angle values (30, 45, 60 degrees) are the only ones used, and answers are left in surd form (e.g., 20 sqrt(3) m) unless decimal approximation is specifically requested.
What is Applications of Trigonometry - Heights & Distances, Elevation, Depression?
The angle formed between the horizontal line of sight and the line of sight to an object above the horizontal level.
- Observer looks UP from the horizontal.
- The angle is measured from the horizontal to the line of sight.
The angle formed between the horizontal line of sight and the line of sight to an object below the horizontal level.
- Observer looks DOWN from the horizontal.
- The angle is measured from the horizontal downward to the line of sight.
Key fact:
Angle of elevation from A to B = Angle of depression from B to A
(alternate interior angles, since the horizontals at A and B are parallel)
Applications of Trigonometry Formula
Key Trigonometric Ratios Used:
tan theta = Opposite/Adjacent = Height/Distance
sin theta = Opposite/Hypotenuse
cos theta = Adjacent/Hypotenuse
Most commonly used ratio:
- tan theta is used most often because problems typically involve a height (opposite) and a horizontal distance (adjacent).
Standard values used in problems:
| Angle | tan | sin | cos |
|---|---|---|---|
| 30 degrees | 1/sqrt(3) | 1/2 | sqrt(3)/2 |
| 45 degrees | 1 | 1/sqrt(2) | 1/sqrt(2) |
| 60 degrees | sqrt(3) | sqrt(3)/2 | 1/2 |
Derivation and Proof
Setting Up Height and Distance Problems:
- Draw a clear diagram showing the observer, the object, and the horizontal.
- Mark the angle of elevation or depression at the observer's position.
- Identify the right triangle formed by:
- The vertical height (opposite side)
- The horizontal distance (adjacent side)
- The line of sight (hypotenuse)
- Apply the appropriate trigonometric ratio (usually tan theta).
- Solve for the unknown (height, distance, or angle).
Key principle:
When the observer looks up at angle theta from horizontal, the right triangle has:
- Perpendicular (height of object above observer) = opposite side
- Base (horizontal distance) = adjacent side
- tan theta = height/distance
Important conventions:
- Unless stated otherwise, assume all objects are vertical (towers, poles, buildings) and the ground is horizontal.
- The line of sight is the straight line from the observer's eye to the object being observed.
- The horizontal is the line from the observer's eye parallel to the ground.
- The angle of elevation/depression is always measured from the horizontal, never from the vertical.
- When the observer has a specific height (e.g., a person 1.5 m tall), measurements are from eye level, not from the ground.
Diagram conventions:
- Draw the ground as a horizontal line.
- Draw vertical objects as vertical lines perpendicular to the ground.
- Mark the angle of elevation at the observer's position (ground level or eye level).
- Mark the right angle at the base of the vertical object.
- Label all known lengths and the unknown(s) with variables.
Types and Properties
Types of height and distance problems:
| Problem Type | Setup |
|---|---|
| Finding height from ground | Given: angle of elevation and horizontal distance. Find: height using tan theta = h/d. |
| Finding distance | Given: angle of elevation and height. Find: distance using d = h/tan theta. |
| Two angles from same point | Object's top and bottom at different angles from the same observer — form two right triangles. |
| Two positions, same object | Observer at two positions with different angles — set up two equations and solve. |
| Object on top of another | Tower on a building, flag on a tower — involves angle to the top and bottom of the upper object. |
| Moving towards/away from object | Observer moves and angle changes — two right triangles with a shared height. |
Methods
Method for solving height and distance problems:
- Read carefully — identify what is given and what is to be found.
- Draw a diagram — this is essential. Label all known quantities.
- Mark the right angle(s) — the height is perpendicular to the ground.
- Identify the triangle(s) — each angle of elevation/depression gives one right triangle.
- Choose the ratio — tan theta is used most often (connects height and distance).
- Set up equation(s) — write tan theta = opposite/adjacent.
- Solve — for single-triangle problems, solve directly. For two-triangle problems, solve simultaneously.
Tips:
- Always assume the ground is horizontal and objects are vertical (unless stated otherwise).
- When the problem says "angle of depression from top of a tower," the angle is at the top, measured from the horizontal downward.
- The angle of depression from A to B equals the angle of elevation from B to A.
- Use sqrt(3) = 1.732 for final numerical answers when asked.
Solved Examples
Example 1: Finding the Height of a Tower
Problem: The angle of elevation of the top of a tower from a point on the ground 30 m from its base is 60 degrees. Find the height of the tower.
Solution:
Given:
- Horizontal distance (d) = 30 m
- Angle of elevation = 60 degrees
Using: tan 60 = h/d
- sqrt(3) = h/30
- h = 30 sqrt(3) = 30 x 1.732 = 51.96 m
Answer: Height of the tower = 30 sqrt(3) m (approximately 51.96 m).
Example 2: Finding the Distance from the Base
Problem: A kite is flying at a height of 60 m above the ground. The string makes an angle of 30 degrees with the horizontal. Find the length of the string.
Solution:
Given:
- Height = 60 m
- Angle with horizontal = 30 degrees
Using: sin 30 = height/length of string
- 1/2 = 60/length
- Length = 120 m
Answer: Length of the string = 120 m.
Example 3: Angle of Depression Problem
Problem: From the top of a 45 m high lighthouse, the angle of depression of a ship is 30 degrees. How far is the ship from the base of the lighthouse?
Solution:
Given:
- Height of lighthouse = 45 m
- Angle of depression = 30 degrees
Angle of depression from top = angle of elevation from ship = 30 degrees.
Using: tan 30 = 45/d
- 1/sqrt(3) = 45/d
- d = 45 sqrt(3) = 45 x 1.732 = 77.94 m
Answer: The ship is 45 sqrt(3) m (approximately 77.94 m) from the base.
Example 4: Two Angles of Elevation — Finding Height
Problem: From two points on the ground on the same side of a tower, the angles of elevation of the top are 60 degrees and 30 degrees. If the two points are 40 m apart, find the height of the tower.
Solution:
Let: h = height, d = distance of nearer point from base.
From nearer point: tan 60 = h/d → sqrt(3) = h/d → d = h/sqrt(3)
From farther point: tan 30 = h/(d + 40) → 1/sqrt(3) = h/(d + 40) → d + 40 = h sqrt(3)
Substituting d = h/sqrt(3):
- h/sqrt(3) + 40 = h sqrt(3)
- 40 = h sqrt(3) - h/sqrt(3) = h(sqrt(3) - 1/sqrt(3)) = h(3-1)/sqrt(3) = 2h/sqrt(3)
- h = 40 sqrt(3)/2 = 20 sqrt(3) = 34.64 m
Answer: Height of the tower = 20 sqrt(3) m (approximately 34.64 m).
Example 5: Object on Top of a Building
Problem: A flagstaff stands on the top of a 5 m high building. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60 degrees and 45 degrees respectively. Find the height of the flagstaff.
Solution:
Let: Height of flagstaff = h, distance from building = d.
To bottom of flagstaff (top of building):
- tan 45 = 5/d → 1 = 5/d → d = 5 m
To top of flagstaff:
- tan 60 = (5 + h)/d → sqrt(3) = (5 + h)/5
- 5 sqrt(3) = 5 + h
- h = 5 sqrt(3) - 5 = 5(sqrt(3) - 1) = 5(1.732 - 1) = 5 x 0.732 = 3.66 m
Answer: Height of the flagstaff = 5(sqrt(3) - 1) m (approximately 3.66 m).
Example 6: Width of a River
Problem: 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 45 degrees. On retreating 20 m from the bank, the angle becomes 30 degrees. Find the width of the river.
Solution:
Let: h = height of tree, d = width of river.
From the bank: tan 45 = h/d → h = d
After retreating 20 m: tan 30 = h/(d + 20) → 1/sqrt(3) = h/(d + 20)
Substituting h = d:
- d + 20 = d sqrt(3)
- 20 = d(sqrt(3) - 1)
- d = 20/(sqrt(3) - 1)
- Rationalise: d = 20(sqrt(3) + 1)/((sqrt(3) - 1)(sqrt(3) + 1)) = 20(sqrt(3) + 1)/2 = 10(sqrt(3) + 1)
- d = 10(1.732 + 1) = 10 x 2.732 = 27.32 m
Answer: Width of the river = 10(sqrt(3) + 1) m (approximately 27.32 m).
Example 7: Two Observers on Same Side
Problem: A 1.5 m tall boy is standing at some distance from a 30 m high building. The angle of elevation from his eyes to the top of the building is 60 degrees. Find the distance between the boy and the building.
Solution:
Effective height = 30 - 1.5 = 28.5 m (from the boy's eye level to the top).
Using: tan 60 = 28.5/d
- sqrt(3) = 28.5/d
- d = 28.5/sqrt(3) = 28.5 sqrt(3)/3 = 9.5 sqrt(3) = 16.45 m
Answer: Distance = 9.5 sqrt(3) m (approximately 16.45 m).
Example 8: Angles of Depression from Top of a Building
Problem: From the top of a 120 m high building, the angles of depression of the top and bottom of a tower are 30 degrees and 60 degrees respectively. Find the height of the tower.
Solution:
Let: Height of tower = h, horizontal distance = d.
To bottom of tower: tan 60 = 120/d → d = 120/sqrt(3) = 40 sqrt(3)
To top of tower: tan 30 = (120 - h)/d → 1/sqrt(3) = (120 - h)/(40 sqrt(3))
- 40 sqrt(3)/sqrt(3) = 120 - h
- 40 = 120 - h
- h = 80 m
Answer: Height of the tower = 80 m.
Example 9: Shadow Problem
Problem: The shadow of a tower, when the angle of elevation of the sun is 45 degrees, is found to be 10 m longer than when it was 60 degrees. Find the height of the tower.
Solution:
Let: h = height, shadow at 60 degrees = d.
When angle = 60: tan 60 = h/d → d = h/sqrt(3)
When angle = 45: tan 45 = h/(d + 10) → h = d + 10
Substituting:
- h = h/sqrt(3) + 10
- h - h/sqrt(3) = 10
- h(1 - 1/sqrt(3)) = 10
- h(sqrt(3) - 1)/sqrt(3) = 10
- h = 10 sqrt(3)/(sqrt(3) - 1)
- Rationalise: h = 10 sqrt(3)(sqrt(3) + 1)/2 = 10(3 + sqrt(3))/2 = 5(3 + sqrt(3))
- h = 5(3 + 1.732) = 5 x 4.732 = 23.66 m
Answer: Height = 5(3 + sqrt(3)) m (approximately 23.66 m).
Example 10: Speed and Distance Problem
Problem: An airplane is flying at a height of 3000 m. The angle of elevation from a point on the ground changes from 60 degrees to 30 degrees in 15 seconds. Find the speed of the airplane.
Solution:
Let: d1 = horizontal distance at 60 degrees, d2 = distance at 30 degrees.
- tan 60 = 3000/d1 → d1 = 3000/sqrt(3) = 1000 sqrt(3)
- tan 30 = 3000/d2 → d2 = 3000 sqrt(3)
Distance covered = d2 - d1 = 3000 sqrt(3) - 1000 sqrt(3) = 2000 sqrt(3) m
Speed = distance/time = 2000 sqrt(3)/15 = 2000 x 1.732/15 = 230.9 m/s
Converting: 230.9 x 3600/1000 = 831.4 km/h
Answer: Speed = 2000 sqrt(3)/15 m/s (approximately 831 km/h).
Example 11: Two Buildings Problem
Problem: Two buildings are on opposite sides of a road 15 m wide. From a point on the road between the buildings, the angles of elevation of the tops are 60 degrees and 45 degrees. Find the heights of the buildings.
Solution:
Let: Point P is on the road. Let PA = d (distance to first building), so PB = 15 - d.
Building 1: tan 60 = h1/d → h1 = d sqrt(3)
Building 2: tan 45 = h2/(15-d) → h2 = 15 - d
Without additional information, the heights depend on the position of P. If P is the point where the angles are as stated, we need one more condition. Often the problem states that the total height or a ratio is given.
If the point is such that d = 15 - d (midpoint), then d = 7.5:
- h1 = 7.5 sqrt(3) = 12.99 m
- h2 = 7.5 m
Answer: Heights depend on position. At the midpoint: h1 = 7.5 sqrt(3) m, h2 = 7.5 m.
Real-World Applications
Real-world applications of trigonometry in heights and distances:
- Surveying: Measuring heights of mountains, buildings, and towers without climbing them.
- Navigation: Ships and aircraft use angles of elevation and depression for position determination.
- Construction: Calculating roof slopes, ramp angles, and support beam lengths.
- Astronomy: Measuring distances to celestial objects using parallax angles.
- Military: Calculating distances to targets using observation angles.
- Geography: Measuring widths of rivers, valleys, and canyons from one side.
- Architecture: Designing buildings with specific shadow requirements (sunlight studies).
- Telecommunications: Calculating cell tower coverage based on tower height and signal angle.
Key Points to Remember
- Angle of elevation: Angle above the horizontal when looking UP at an object.
- Angle of depression: Angle below the horizontal when looking DOWN at an object.
- Angle of elevation from A to B = Angle of depression from B to A (alternate interior angles).
- tan theta = height/distance is the most commonly used ratio in these problems.
- Always draw a diagram before solving.
- Identify the right triangle(s) in the problem.
- For two-angle problems: set up two equations with two unknowns and solve simultaneously.
- Standard values: tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3).
- Use sqrt(3) = 1.732 for decimal approximation.
- If the observer has a height (e.g., a boy 1.5 m tall), subtract the observer's height from the object's height to get the effective height.
Practice Problems
- The angle of elevation of the top of a tower from a point 50 m away from its base is 45 degrees. Find the height of the tower.
- From the top of a 60 m cliff, the angle of depression of a boat is 30 degrees. Find the distance of the boat from the base of the cliff.
- A ladder 10 m long reaches a window 8 m above the ground. Find the angle the ladder makes with the ground.
- The shadow of a 15 m high pole is 5 sqrt(3) m. Find the angle of elevation of the sun.
- From two points 80 m apart on the same side of a tower, the angles of elevation are 30 degrees and 60 degrees. Find the height of the tower.
- A 7 m tall flagstaff is on top of a building. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45 degrees and 30 degrees. Find the height of the building.
- From the top of a 100 m tower, the angles of depression of the top and bottom of a building are 45 degrees and 60 degrees. Find the height of the building.
- An airplane is flying at a height of 5000 m. The angle of elevation from a point on the ground is 30 degrees. Find the horizontal distance of the airplane from that point.
Frequently Asked Questions
Q1. What is the angle of elevation?
The angle formed between the horizontal line of sight and the line of sight upward to an object. It is measured at the observer's position when looking UP.
Q2. What is the angle of depression?
The angle formed between the horizontal line of sight and the line of sight downward to an object. It is measured at the observer's position when looking DOWN.
Q3. Why is tan theta used most often?
Because most problems involve a vertical height (opposite side) and a horizontal distance (adjacent side). tan theta = opposite/adjacent directly connects height and distance.
Q4. What is the relationship between angle of elevation and angle of depression?
The angle of elevation from point A to point B equals the angle of depression from point B to point A. This is because the horizontal lines at A and B are parallel, making these alternate interior angles.
Q5. How do you handle two-angle problems?
Set up two right triangles sharing a common height. Write two equations using tan for each angle. Then solve the system of equations by substitution or elimination.
Q6. Do I always need to draw a diagram?
Yes. Drawing a clear, labelled diagram is the first and most important step. Without it, setting up the correct trigonometric equation is very difficult.
Q7. What if the observer is not at ground level?
Subtract the observer's height from the object's height to get the effective height for the right triangle. For example, a 1.5 m tall person looking at a 20 m building has an effective height of 18.5 m.
Q8. Are these problems important for board exams?
Yes. Height and distance problems are among the most commonly asked questions, typically worth 4-5 marks. They require drawing a diagram, setting up the equation, and solving with standard trigonometric values.
