Courses
By Shailendra Singh
|
Updated on 12 Nov 2025, 15:36 IST
Trigonometry finds extensive practical applications in real-world scenarios where direct measurement is challenging or impossible. From determining the height of towering buildings and mountains to calculating distances between celestial objects, trigonometry serves as an indispensable mathematical tool in fields like geography, astronomy, navigation, architecture, and surveying.
In this chapter, we explore how trigonometric ratios help solve problems involving heights and distances through indirect methods using right triangles, angles of elevation, and angles of depression.
The line of sight is the straight line drawn from the observer's eye to the point being viewed on an object. This imaginary line forms the basis for measuring angles of elevation and depression.
A horizontal line is an imaginary line parallel to the ground, passing through the observer's eye level. All angular measurements are made with reference to this horizontal line.
The angle of elevation is the angle formed between the horizontal line and the line of sight when an observer looks upward at an object positioned above eye level.
Key Points:
Loading PDF...
Example: When you stand on the ground and look up at the top of a building, the angle your line of sight makes with the horizontal is the angle of elevation.
The angle of depression is the angle formed between the horizontal line and the line of sight when an observer looks downward at an object positioned below eye level.
Key Points:
Example: When you stand on top of a lighthouse and look down at a ship in the sea, the angle your line of sight makes with the horizontal is the angle of depression.
Critical Observation: The angle of depression of object P as seen from observer O is equal to the angle of elevation of observer O as seen from object P. This is due to the property of alternate interior angles formed when parallel horizontal lines are cut by the line of sight.
| Trigonometric Ratio | Formula | Explanation |
| Sine (sin θ) | sin θ = Perpendicular / Hypotenuse | Ratio of opposite side to hypotenuse |
| Cosine (cos θ) | cos θ = Base / Hypotenuse | Ratio of adjacent side to hypotenuse |
| Tangent (tan θ) | tan θ = Perpendicular / Base | Ratio of opposite side to adjacent side |
| Cosecant (cosec θ) | cosec θ = 1 / sin θ | Reciprocal of sine |
| Secant (sec θ) | sec θ = 1 / cos θ | Reciprocal of cosine |
| Cotangent (cot θ) | cot θ = 1 / tan θ | Reciprocal of tangent |
| Angle (θ) | sin θ | cos θ | tan θ | cot θ | sec θ | cosec θ |
| 0° | 0 | 1 | 0 | ∞ | 1 | ∞ |
| 30° | 1/2 | √3/2 | 1/√3 | √3 | 2/√3 | 2 |
| 45° | 1/√2 | 1/√2 | 1 | 1 | √2 | √2 |
| 60° | √3/2 | 1/2 | √3 | 1/√3 | 2 | 2/√3 |
| 90° | 1 | 0 | ∞ | 0 | ∞ | 1 |
| Identity | Explanation |
| sin(90° - θ) = cos θ | Sine of complement equals cosine |
| cos(90° - θ) = sin θ | Cosine of complement equals sine |
| tan(90° - θ) = cot θ | Tangent of complement equals cotangent |
Step 1: Draw a Clear Diagram
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Step 2: Identify the Right Triangle
Step 3: Select Appropriate Trigonometric Ratio Use the formula: Required side / Given side = Appropriate T-ratio of the given angle
Step 4: Form the Equation Set up the trigonometric equation based on the chosen ratio.
Step 5: Solve for Unknown Use standard trigonometric values to calculate the required measurement.
Step 6: Verify Units Ensure your answer is in the correct units and makes practical sense.
The approach is similar to angle of elevation problems with one key difference:
Strategy: Convert the angle of depression to an angle of elevation using the alternate angle property. The angle of depression from the top equals the angle of elevation from the bottom.
Steps:
Problem: An observer 1.5 m tall is standing 28.5 m away from a tower that is 30 m high. Determine the angle of elevation of the top of the tower from his eye.
Solution:
Let AB = height of tower = 30 m
CD = height of observer = 1.5 m
AC = horizontal distance = 28.5 m
Through point D (observer's eye), draw DE || CA.
Then:
In right triangle BDE:
tan θ = BE / DE = 28.5 / 28.5 = 1
tan θ = 1 = tan 45°
Therefore, θ = 45°
Answer: The angle of elevation is 45°.
Problem: A vertical post casts a shadow 21 m long when the altitude (angle of elevation) of the sun is 30°. Find:
Solution:
(a) Finding height when angle = 30° and shadow = 21 m:
Let AB = h (height of post)
BC = 21 m (shadow length)
∠ACB = 30°
In right triangle ABC:
tan 30° = AB / BC
1/√3 = h / 21
h = 21/√3 = 21/√3 × √3/√3 = 21√3/3 = 7√3 m
(b) Finding shadow length when angle = 60°:
Now ∠ACB = 60°, AB = 7√3 m, BC = x m
tan 60° = AB / BC
√3 = 7√3 / x
x = 7√3 / √3 = 7 m
(c) Finding angle when shadow = 7√3 m:
AB = 7√3 m, BC = 7√3 m
tan θ = AB / BC = 7√3 / 7√3 = 1
tan θ = tan 45°
Therefore, θ = 45°
Problem: A 1.6 m tall girl stands 3.2 m from a lamp-post and casts a 4.8 m shadow. Find the height of the lamp-post using:
Solution:
Let PQ = h (height of lamp-post)
AB = 1.6 m (girl's height)
AE = BQ = 3.2 m
BC = 4.8 m (shadow length)
(i) Using Trigonometry:
In right triangle ABC:
tan θ = AB / BC = 1.6 / 4.8 = 1/3
In right triangle AEP:
PE = PQ - EQ = h - 1.6
AE = 3.2 m
tan θ = PE / AE
1/3 = (h - 1.6) / 3.2
3.2 = 3(h - 1.6)
3.2 = 3h - 4.8
3h = 8
h = 8/3 = 2⅔ m
(ii) Using Similar Triangles:
△ACB ~ △PCQ (AA similarity)
Therefore: AC / PC = CB / CQ = AB / PQ
(4.8) / 8 = 1.6 / h
4.8h = 8 × 1.6 = 12.8
h = 12.8 / 4.8 = 8/3 = 2⅔ m
Problem: A captain flying at 1000 m altitude sights two ships at angles of depression 60° and 30°. If one ship is directly behind the other, find the distance between them.
Solution:
Let A = position of aeroplane
AB = 1000 m (altitude)
P and Q = positions of two ships
∠PAB = 60°, ∠QAB = 30° (angles of depression)
These equal the angles of elevation from ships:
∠APB = 60°, ∠AQB = 30°
In right triangle ABP:
tan 60° = AB / BP
√3 = 1000 / BP
BP = 1000/√3 = 1000√3/3 m
In right triangle ABQ:
tan 30° = AB / BQ
1/√3 = 1000 / BQ
BQ = 1000√3 m
Distance between ships:
PQ = BQ - BP = 1000√3 - 1000√3/3
PQ = 1000√3(1 - 1/3) = 1000√3 × 2/3
PQ = 2000√3/3 m ≈ 2309.3 m
Problem: A 1.5 m tall boy standing at some distance from a 30 m building observes that the angle of elevation increases from 30° to 60° as he walks toward it. Find the distance he walked.
Solution:
Let the boy initially be at S, then move to T.
PR = PQ - RQ = 30 - 1.5 = 28.5 m = 57/2 m
From position S (angle 30°):
tan 30° = PR / AR
1/√3 = (57/2) / AR
AR = (57/2) × √3 = 57√3/2 m
From position T (angle 60°):
tan 60° = PR / BR
√3 = (57/2) / BR
BR = (57/2) ÷ √3 = 57/(2√3) = 19√3/2 m
Distance walked:
ST = AR - BR = 57√3/2 - 19√3/2
ST = (57√3 - 19√3)/2 = 38√3/2
ST = 19√3 m ≈ 32.9 m
Application: Ships and aircraft use trigonometry to:
Example: A ship needs to calculate its distance from a lighthouse. By measuring the angle of elevation to the lighthouse top and knowing its height, navigators use trigonometric ratios to find the exact distance.
Application: Surveyors employ trigonometry to:
Method: Using theodolites (instruments measuring angles), surveyors form triangles and apply trigonometric principles to calculate distances without physically measuring them.
Application: Astronomers use trigonometry to:
Technique: Parallax method uses angles of observation from different points to calculate vast astronomical distances.
Application: Architects and engineers use trigonometry to:
Application: Geographers apply trigonometry to:
In a right triangle with angle θ:
sin θ = Perpendicular / Hypotenuse = P / H
cos θ = Base / Hypotenuse = B / H
Therefore:
tan θ = sin θ / cos θ = (P/H) / (B/H) = P/B = Perpendicular / Base
This shows that tangent can be derived from sine and cosine.
Consider a right triangle with angles 90°, θ, and (90° - θ).
For angle θ:
For angle (90° - θ):
Therefore:
sin(90° - θ) = Opposite/Hypotenuse for (90° - θ) = B/H = cos θ
cos(90° - θ) = Adjacent/Hypotenuse for (90° - θ) = P/H = sin θ
tan(90° - θ) = Opposite/Adjacent for (90° - θ) = B/P = cot θ
Applications of Trigonometry typically carries 4-6 marks in CBSE Class 10 Mathematics exam, usually as:
High Priority:
Medium Priority:
Q: What is the difference between angle of elevation and angle of depression?
A: Angle of elevation is measured upward from the horizontal when looking at an object above eye level. Angle of depression is measured downward from the horizontal when looking at an object below eye level. They are alternate angles and equal when measured from related positions.
Q: Why do we use tan θ most frequently in height-distance problems?
A: Tangent ratio (tan θ = perpendicular/base) directly relates the two quantities we typically deal with—vertical height and horizontal distance—without involving the hypotenuse, making calculations simpler.
Q: How do I know which trigonometric ratio to use?
A: Identify what is given and what is required:
Q: Can the angle of elevation ever be greater than 90°?
A: No. By definition, angle of elevation is measured from the horizontal upward to the line of sight. The maximum theoretical value is 90° (looking straight up), but practical problems use angles less than 90°.
Q: Do I always need to rationalize the denominator?
A: Yes, for CBSE exams, it's best practice to express answers with rationalized denominators. For example, write 5√3/3 instead of 5/√3.
Applications of Trigonometry is a highly practical chapter that bridges pure mathematics with real-world problem-solving. Mastery of angle of elevation, angle of depression, and height-distance calculations opens doors to understanding how mathematics applies to navigation, surveying, architecture, and astronomy.
Key to success:
With consistent practice of varied problem types and attention to fundamental concepts, you'll develop strong proficiency in applying trigonometry to real-life situations.
Related Resources:
No courses found
The angle of elevation is the angle formed between a horizontal line drawn through an observer's eye and the line of sight when looking upward at an object. For example, when you look up at the top of a building from ground level, the angle your line of sight makes with the horizontal ground is the angle of elevation. This concept is crucial for calculating heights and distances of objects like towers, buildings, and trees without direct measurement.
The angle of depression is the angle formed between a horizontal line through an observer's eye and the line of sight when looking downward at an object below the horizontal level. The key difference is direction: angle of elevation involves looking upward, while angle of depression involves looking downward. Interestingly, the angle of depression from point A to point B equals the angle of elevation from point B to point A due to alternate interior angles formed by parallel horizontal lines.
The line of sight is the straight line drawn from the observer's eye to the object being viewed. It's the direct visual path between the observer and the target object. This line, combined with the horizontal line through the observer's eye, forms either the angle of elevation or angle of depression, which are essential for solving height and distance problems in real-world applications.
To find the height of a tower using trigonometry, you need to know the horizontal distance from the tower's base and the angle of elevation to its top. Using the tangent ratio (tan θ = height/base), you can calculate the tower's height. For instance, if you're standing 50 meters from a tower and the angle of elevation is 60°, the height would be 50 × tan(60°) = 50 × √3 ≈ 86.6 meters. This method is particularly useful when direct measurement is impractical or impossible.
Trigonometry is essential in surveying and navigation because it allows professionals to measure distances and heights indirectly. Surveyors use trigonometric principles to create accurate maps, determine land boundaries, and establish positions relative to longitudes and latitudes. In navigation, especially in astronomy and maritime travel, trigonometry helps determine a vessel's position, plot courses, and calculate distances between locations. These applications have been fundamental to exploration and mapping throughout history.
When dealing with two angles of elevation from the same observation point, you typically create two right triangles sharing a common side (usually the height you're solving for). Set up separate trigonometric equations for each triangle, then solve simultaneously. For example, if the angles of elevation to the top and bottom of a flagpole on a building are given, you can find both the building's height and the flagpole's length by subtracting one calculated height from the other.
When the observer's height is not explicitly given in a trigonometry problem, you should assume the observer is standing at ground level with negligible height. This simplifies calculations by treating the angle of elevation as measured directly from the ground. However, if the observer's height is provided (for instance, 1.5 meters tall), you must account for this by adjusting your final answer the total height equals the calculated height plus the observer's eye level.
To find distances between two objects using angles of depression, observe both objects from an elevated point (like a lighthouse or airplane). The angles of depression to each object, combined with the known height of the observation point, allow you to calculate the horizontal distance to each object using trigonometric ratios. The distance between the two objects is simply the difference between these two horizontal distances. This technique is commonly used in maritime navigation and aerial surveying.
For practical height and distance problems, you should memorize these key values:
These values for 30°, 45°, and 60° angles appear frequently in real-world applications and standardized problems.
Choose your trigonometric ratio based on what information you have:
The fundamental rule is: Required side / Given side = appropriate trigonometric ratio of the given angle.
When two angles of elevation from different points are complementary (they add up to 90°), one angle is θ and the other is (90° - θ). This creates a special relationship because tan(θ) × tan(90° - θ) = tan(θ) × cot(θ) = 1. If you're measuring a tower's height from two points at distances of 4m and 9m with complementary angles, you can multiply the two tangent equations to get: (h/4) × (h/9) = 1, giving h² = 36, so h = 6 meters.
Shadow length changes with the sun's altitude (angle of elevation). Using the formula: shadow length = height / tan(altitude angle). For a 7-meter post:
As the sun rises higher (larger angle), shadows become shorter.
When an observer moves toward an object and the angle of elevation changes, you can determine the distance traveled. Set up two separate triangles for the initial and final positions, both sharing the object's height as a common side. Calculate the horizontal distance for each position, then subtract to find the distance traveled. For example, if a boy walks toward a building and his angle of elevation changes from 30° to 60°, you can calculate both distances from the building and find the difference.
When a tree breaks and the broken part touches the ground while still attached, it forms a right triangle. You need the angle the broken part makes with the ground and the horizontal distance from the base to where the top touches. The standing part equals (distance × tan(angle)), and the broken part equals (distance / cos(angle)). The total original height is the sum of both parts. For instance, with an 8-meter distance and 30° angle: standing part = 8/√3 meters, broken part = 16/√3 meters, total = 24/√3 = 8√3 meters.
When standing between two poles of equal height with different angles of elevation, you can find both the poles' height and your position. Let one distance be x and the other (80 - x) if the total distance is 80m.
With angles of 60° and 30°, set up: h = x×tan(60°) = x√3 and h = (80-x)×tan(30°) = (80-x)/√3. Solving these simultaneously: x√3 = (80-x)/√3, which gives 3x = 80-x, so 4x = 80, thus x = 20 meters. The height would be h = 20√3 ≈ 34.64 meters.
Common mistakes include:
Always draw a diagram first and clearly label all known values and the quantity you're solving for.
Bearing is the direction of one point relative to another, measured in degrees clockwise from north. A bearing is expressed as a three-digit number from 000° to 360°, where north is 000°/360°, east is 090°, south is 180°, and west is 270°. In navigation and surveying, bearings combined with distances and trigonometry help determine exact positions, plot courses, and calculate the coordinates of landmarks. For instance, if a ship travels on a bearing of 045° (northeast) for a certain distance, you can use trigonometry to find how far north and how far east it has traveled.
For mathematical convenience and practical purposes, objects like towers, trees, buildings, and mountains are considered as straight vertical lines in trigonometry problems. This simplification allows us to form clear right triangles and apply trigonometric ratios consistently. While real objects have width and may not be perfectly vertical, this approximation introduces negligible error for most practical applications, especially when dealing with tall structures where height is the primary dimension of interest. This assumption is one of the fundamental conventions in applied trigonometry.