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Updated on 23 Jul 2026, 14:32 IST
Some Applications of Trigonometry explains how trigonometric ratios are used to calculate heights and distances that cannot be measured directly. These Some Applications of Trigonometry Class 10 notes PDF cover line of sight, angles of elevation and depression, diagram construction, formula selection, solved examples, common mistakes, important questions and board-exam revision.
Also Check: Introduction to Trigonometry Class 10 Notes
Some Applications of Trigonometry uses the properties of right-angled triangles to find unknown heights, horizontal distances and lines of sight.
In the previous trigonometry chapter, you learn ratios such as sine, cosine and tangent. In this chapter, you apply those ratios to situations involving towers, buildings, trees, poles, shadows, ships and observation points.
The NCERT chapter is titled “Some Applications of Trigonometry” and mainly covers heights and distances.
The Applications of Trigonometry Class 10 notes PDF should include complete explanations, labelled diagrams, formulas, worked examples and practice answers in one printable resource.
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Heights and distances: Problems in which trigonometric ratios are used to calculate the vertical height of an object or the horizontal distance between two points.
A height may be difficult or unsafe to measure directly. For example, measuring a tall tower with a tape is impractical. Trigonometry allows us to calculate its height using:
Trigonometry can be used to estimate:
Surveyors, architects, engineers and navigators use more advanced forms of the same mathematical ideas.
For exam preparation, students should follow the scope stated in the latest official CBSE curriculum rather than treating every online trigonometry problem as syllabus material.
The current curriculum limits heights-and-distances applications to:
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
The complete Trigonometry unit carries 12 marks out of the 80-mark theory paper. This weightage covers the full unit, including trigonometric ratios, identities and applications; it does not mean this chapter alone carries 12 marks.
Official CBSE Class 10 sample papers for Mathematics Standard and Mathematics Basic contain 38 compulsory questions arranged in five sections.
The main terms in this chapter describe the observer, the object being viewed and the lines and angles connecting them.
| Term | Definition |
| Observer | The person or point from which an object is viewed |
| Eye level | The horizontal level passing through the observer’s eyes |
| Horizontal line | A line parallel to level ground |
| Line of sight | The straight line joining the observer’s eye to the point being viewed |
| Angle of elevation | The angle between the horizontal and an upward line of sight |
| Angle of depression | The angle between the horizontal and a downward line of sight |
| Vertical height | The perpendicular distance from the base to the top of an object |
| Horizontal distance | The distance measured parallel to level ground |
| Foot of an object | The point where a vertical object meets the ground |
| Observation point | The position from which an observer views an object |
| Reference angle | The angle used to identify the opposite and adjacent sides |
| Hypotenuse | The side opposite the right angle; it is the longest side of a right triangle |
The line of sight is the straight line joining the observer’s eye to the point on the object that the observer is viewing.
In a right-triangle model, the line of sight may form the hypotenuse. It is not automatically the same as the horizontal distance.
The angle of elevation is the angle measured upward from the observer’s horizontal line to the line of sight.
For example, when a student standing on the ground looks at the top of a tower, the angle between the eye-level horizontal and the upward line of sight is the angle of elevation.
The angle of depression is the angle measured downward from the observer’s horizontal line to the line of sight.
For example, when a person at the top of a lighthouse looks down at a ship, the angle between the horizontal through the observer and the downward line of sight is the angle of depression.
The difference between elevation and depression is the direction in which the observer looks from the horizontal.
| Feature | Angle of elevation | Angle of depression |
| Viewing direction | Upward | Downward |
| Starting line | Observer’s horizontal | Observer’s horizontal |
| Observer is usually | Below the viewed point | Above the viewed point |
| Common example | Looking at a tower top | Looking down from a lighthouse |
| Angle location | At the lower observation point | At the upper observation point |
An angle of depression can equal the corresponding angle of elevation because the two horizontal lines are parallel and the line of sight acts as a transversal.
The equal angles are alternate interior angles.
This relationship allows an angle marked outside the right triangle at the upper observer to be transferred to an equal angle inside the working triangle.
A correct diagram turns a word problem into one or two right-angled triangles whose sides and angles can be analysed.
Use this diagram-first method before selecting a formula.
Place an angle of elevation at the observer, between the horizontal line and the upward line of sight.
Correct setup:
Incorrect setup:
Place an angle of depression at the upper observer, between the eye-level horizontal and the downward line of sight.
A common mistake is to measure the angle from a vertical wall. The depression angle is always measured from a horizontal line.
When the angle is measured from the observer’s eye, the right triangle may calculate only the height above eye level.
Suppose:
Then:
Total height = 15 + 1.5 = 16.5 m
Add the observer’s height only when the diagram shows that the calculated side begins at eye level rather than ground level.
A trigonometry diagram does not usually need to be drawn to scale, but it must show the correct mathematical relationships.
A useful diagram must have:
Do not estimate an answer by measuring the drawing.
| Mistake | Why it causes an error | Correction |
| Angle measured from a vertical line | Elevation and depression are measured from a horizontal | Draw an eye-level horizontal first |
| Angle placed at the object | The angle belongs at the observation point | Mark the observer before drawing the angle |
| Missing right angle | The right-triangle relationship becomes unclear | Mark where the vertical object meets level ground |
| Line of sight treated as ground distance | The line of sight is often the hypotenuse | Label the horizontal and diagonal separately |
| Observer height ignored | The calculated height may be too small | Add the eye height when required |
| Two observation points merged | The distances from the object are different | Use separate variables or expressions |
| Diagram assumed to be to scale | Visual proportions may be misleading | Use only the stated measurements |
Sine, cosine and tangent connect an acute angle in a right triangle with ratios of its side lengths.
For a reference angle θ:
| Ratio | Formula |
| sin θ | Opposite ÷ Hypotenuse |
| cos θ | Adjacent ÷ Hypotenuse |
| tan θ | Opposite ÷ Adjacent |
| cosec θ | Hypotenuse ÷ Opposite |
| sec θ | Hypotenuse ÷ Adjacent |
| cot θ | Adjacent ÷ Opposite |
A useful memory aid is:
SOH: Sine = Opposite/Hypotenuse
CAH: Cosine = Adjacent/Hypotenuse
TOA: Tangent = Opposite/Adjacent
Choose the ratio containing the side you know and the side you need to calculate.
| Side known | Side required | Ratio to consider |
| Adjacent | Opposite | Tangent |
| Opposite | Adjacent | Tangent |
| Hypotenuse | Opposite | Sine |
| Opposite | Hypotenuse | Sine |
| Hypotenuse | Adjacent | Cosine |
| Adjacent | Hypotenuse | Cosine |
Do not select a ratio only because it appears frequently. First identify the two sides involved.
Tangent is used often because many heights-and-distances problems connect a vertical height with a horizontal ground distance.
For a tower problem:
Sine or cosine is more suitable when the line of sight or another hypotenuse is given.
The hypotenuse is always opposite the 90° angle. The opposite and adjacent sides depend on the selected acute reference angle.
Hypotenuse: The longest side, opposite the right angle.
Opposite side: The side directly across from the reference angle.
Adjacent side: The non-hypotenuse side touching the reference angle.
If the reference angle changes, the opposite and adjacent sides may exchange roles. The hypotenuse does not change.
Example:
Therefore:
tan θ = h/20
Standard trigonometric values allow exact answers to be calculated without a calculator.
| θ | 0° | 30° | 45° | 60° | 90° |
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | Not defined |
| cosec θ | Not defined | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | Not defined |
| cot θ | Not defined | √3 | 1 | 1/√3 | 0 |
For heights-and-distances applications in the current syllabus, focus especially on 30°, 45° and 60°.
Also Check: Class 10 Maths formula sheet
| Angle | tan value | Useful observation |
| 30° | 1/√3 | Height is smaller than the adjacent distance |
| 45° | 1 | Height equals the adjacent distance |
| 60° | √3 | Height is greater than the adjacent distance |
These observations can help you check whether an answer is reasonable.
Keep √3 in exact form until the final step unless the question asks for an approximate decimal answer.
Example:
tan 30° = 1/√3
If:
1/√3 = h/12
Then:
h = 12/√3
Rationalising:
h = (12√3)/3 = 4√3 m
Using √3 ≈ 1.732:
h ≈ 6.93 m
Round only at the final step and follow any accuracy instruction given in the question.
If no instruction is provided:
Rounding early can produce an inaccurate final answer, especially in multi-step problems.
Every heights-and-distances question can be solved by drawing the model, choosing the correct ratio and solving the resulting equation.
Use this seven-step method:
Question:
A student stands 20 m from the foot of a vertical tower. The angle of elevation of the top of the tower is 45°. Find the tower’s height.
Given:
The height is opposite the angle and the 20 m distance is adjacent.
tan 45° = h/20
Since tan 45° = 1:
1 = h/20
h = 20 m
Answer: The tower is 20 m high.
Reasonableness check: At 45°, the opposite and adjacent sides are equal.
Question:
The angle of elevation of the top of a building from a point on level ground is 30°. If the building is 10√3 m high, find the horizontal distance from the point to the building.
Given:
tan 30° = height/distance
1/√3 = 10√3/d
d = 10√3 × √3
d = 30 m
Answer: The observation point is 30 m from the building.
Question:
A student whose eye level is 1.5 m above the ground observes the top of a building at an angle of elevation of 45°. The student is 18 m from the building. Find the building’s total height.
Let the height above the student’s eye level be h.
tan 45° = h/18
1 = h/18
h = 18 m
Total building height:
18 + 1.5 = 19.5 m
Answer: The building is 19.5 m high.
Common mistake: Writing 18 m as the final answer and forgetting the student’s eye height.
Question:
From the top of a 30 m lighthouse, the angle of depression of a ship is 45°. Find the horizontal distance of the ship from the foot of the lighthouse.
The angle of depression equals the corresponding angle of elevation from the ship.
Let the horizontal distance be d.
tan 45° = 30/d
1 = 30/d
d = 30 m
Answer: The ship is 30 m from the foot of the lighthouse.
Applications-of-trigonometry questions can be classified by the number of observation points, number of right triangles and type of unknown measurement.
This is the simplest question type and usually forms one right triangle.
Typical wording:
Common objects:
Use tangent when the question gives a horizontal distance and asks for vertical height.
Formula:
tan θ = height/distance
Therefore:
height = distance × tan θ
Use tangent when the vertical height is known and the horizontal distance is required.
Formula:
tan θ = height/distance
Therefore:
distance = height/tan θ
A shadow problem forms a right triangle using the vertical object, the horizontal shadow and the sun’s ray.
Question:
A vertical pole is 6 m high. If the angle of elevation of the sun is 60°, find the pole’s shadow length.
Let the shadow length be x.
tan 60° = 6/x
√3 = 6/x
x = 6/√3 = 2√3 m
Answer: The shadow is 2√3 m long.
As the sun’s elevation angle increases, the shadow becomes shorter for the same object.
A kite string or a straight line from an observer to a balloon may represent the hypotenuse.
If the string length and angle are given, sine may connect the vertical height with the hypotenuse:
sin θ = vertical height/string length
Check whether the question also gives the height of the observer’s hand above the ground.
These questions contain two vertical measurements:
Total object height = calculated height + observer height
Draw a horizontal line from the observer’s eye to the object to see which part of the total height belongs to the right triangle.
A two-observation-point problem creates two right triangles sharing the same object height.
Typical wording:
Use one tangent equation for each triangle.
Question:
The angle of elevation of the top of a tower from point A is 30°. After moving 20 m closer to the tower to point B, the angle becomes 60°. Find the height of the tower.
Let:
From point B:
tan 60° = h/x
√3 = h/x
h = x√3
From point A:
tan 30° = h/(x + 20)
1/√3 = h/(x + 20)
h = (x + 20)/√3
Equate the two expressions for h:
x√3 = (x + 20)/√3
3x = x + 20
2x = 20
x = 10
Therefore:
h = x√3 = 10√3 m
Answer: The tower is 10√3 m high.
Common mistake: Treating both horizontal distances as x instead of x and x + 20.
When the top of one object is viewed from two positions, form a separate tangent equation from each position.
The basic method is:
A problem may include an observer above ground looking:
Draw the observer’s horizontal line first. Then form a separate triangle for each line of sight.
Do not add the two angles unless the geometry of the question specifically requires it.
Separate the vertical structure into:
An angle to the top of the building forms one triangle. An angle to the top of the tower forms a second triangle. Both may share the same horizontal distance.
When two viewed points lie on the same vertical structure, calculate their heights from the same horizontal distance and subtract if the question asks for the vertical distance between them.
Example:
A river-width problem uses a measurable baseline and an angle of sight to calculate a distance that cannot be measured directly.
Only use a simple right-triangle method when the given points form a perpendicular relationship. More advanced surveying configurations may be beyond the Class 10 syllabus.
Solved examples should show the diagram, selected ratio, equation and final answer rather than providing answers alone.
A person stands 15 m from a tree. The angle of elevation of the top is 30°. Find the tree’s height.
tan 30° = h/15
1/√3 = h/15
h = 15/√3 = 5√3 m
Answer: 5√3 m
A person stands 12 m from a tower. The angle of elevation is 60°. Find its height.
tan 60° = h/12
√3 = h/12
h = 12√3 m
Answer: 12√3 m
The top of a 25 m building is seen at an elevation angle of 45°. Find the observer’s horizontal distance from the building.
tan 45° = 25/d
1 = 25/d
d = 25 m
Answer: 25 m
A tree is 8√3 m tall. Find its shadow length when the sun’s elevation is 60°.
tan 60° = height/shadow
√3 = 8√3/x
x = 8 m
Answer: 8 m
A point on the ground is 10 m from the foot of a pole. The angle of elevation of the top is 30°. Find the line-of-sight distance.
Let the line-of-sight distance be l.
cos 30° = adjacent/hypotenuse
√3/2 = 10/l
l√3 = 20
l = 20/√3 = 20√3/3 m
Answer: 20√3/3 m
A person standing on a 20 m building observes a car at an angle of depression of 45°. Find the car’s horizontal distance from the building.
The corresponding angle of elevation is 45°.
tan 45° = 20/d
1 = 20/d
d = 20 m
Answer: 20 m
A student stands 10 m from a pole and observes its top at 60°. The student’s eye height is 1.4 m. Find the pole’s total height.
Height above eye level:
tan 60° = h/10
√3 = h/10
h = 10√3 m
Total height:
10√3 + 1.4 m
Using √3 ≈ 1.732:
Total height ≈ 17.32 + 1.4 = 18.72 m
Answer: Approximately 18.72 m
The angle of elevation of the top of a tower is 30° from point A and 45° from a nearer point B. The tower is 20 m high. Find the distance AB.
Distance from B:
tan 45° = 20/x
x = 20 m
Distance from A:
tan 30° = 20/y
1/√3 = 20/y
y = 20√3 m
Therefore:
AB = y − x
AB = 20√3 − 20
Answer: 20(√3 − 1) m
From a point 30 m from a building, the angle of elevation of the top of the building is 45°, while the angle of elevation of the top of a tower on the building is 60°. Find the tower’s height.
Building height:
tan 45° = building height/30
Building height = 30 m
Total height:
tan 60° = total height/30
Total height = 30√3 m
Tower height:
30√3 − 30 = 30(√3 − 1) m
Answer: 30(√3 − 1) m
A school wants to estimate the height of a flagpole. A student’s eye level is 1.5 m above the ground. The student stands 12 m from the pole and records an elevation angle of 45°.
Solution:
Answer: The flagpole is 13.5 m high.
Most errors in this chapter come from incorrect diagrams, unsuitable ratio selection or incomplete interpretation of the calculated length.
| Mistake | Why it happens | How to detect it | Correct method |
| Depression measured from vertical | The horizontal through the observer was not drawn | Angle appears between line of sight and wall | Draw the eye-level horizontal |
| Wrong reference angle | Student uses an angle at another vertex | Opposite and adjacent sides do not match the formula | Circle the angle used |
| Observer height omitted | Calculated segment begins at eye level | Answer seems shorter than expected | Add the given eye height |
| Wrong horizontal segment | Two observation points are treated as one | Both equations use the same distance incorrectly | Label each distance separately |
| Sine used instead of tangent | Ratio selected before sides are identified | Hypotenuse is not known or required | Mark known and unknown sides first |
| 30° and 60° values reversed | Standard-value table not checked | A farther point produces a larger angle | Verify tan 30° and tan 60° |
| Early rounding | √3 is replaced too soon | Final answer differs slightly | Keep exact values until the end |
| Missing units | Numerical work is completed without interpretation | Final line contains only a number | Add m, cm or the stated unit |
| Diagram treated as exact scale | Student estimates lengths visually | Calculations conflict with the drawing | Use given values, not visual measurement |
| Total and partial height confused | Two vertical parts are not separated | Wrong part is reported as the answer | Label each vertical segment |
An unexpectedly small height often means that:
A negative physical distance usually indicates an algebra or diagram error.
Check whether:
For a fixed object, moving closer generally increases the angle of elevation.
Use these quick checks:
A complete board-style solution should show the mathematical model, chosen formula, substitution, working and final answer with units.
Official CBSE marking schemes demonstrate that method steps matter, not only the final numerical answer.
For a full solution, write:
Suggested format:
Let the height of the tower be h m.
tan 45° = h/20
1 = h/20
h = 20
Therefore, the height of the tower is 20 m.
A diagram should be drawn whenever it helps translate the situation into a right triangle, even when the question does not explicitly say “draw a diagram.”
Do not assume that every exam awards a fixed mark for a diagram. Marks depend on the particular question and official marking scheme. A correct diagram still reduces the risk of using the wrong angle, side or distance.
The core chapter concepts include the same essential vocabulary, ratios and height-and-distance models, but students should practise questions from the paper type they are taking.
Official CBSE resources provide separate sample question papers and marking schemes for Mathematics Standard, code 041, and Mathematics Basic, code 241.
Prepare for the chapter through several formats:
The official 2025–26 Mathematics Standard and Basic sample papers use five sections and include objective, short-answer, long-answer and case-study formats.
Before the examination, confirm that you can:
Important questions should cover definitions, diagrams, single-triangle calculations, observer height and two-triangle applications.
A survey team wants to estimate the height of a school building. A student stands 20 m from the building. The student’s eye level is 1.6 m, and the elevation angle to the top is 45°.
Answer the following:
Answers:
Only label a question as a previous-year question after verifying its source.
Each verified entry should include:
| Field | Required information |
| Examination year | Exact year |
| Paper type | Mathematics Basic or Standard |
| Marks | Marks assigned |
| Set or code | Where available |
| Question | Exact or accurately transcribed wording |
| Main concept | Elevation, depression, observer height or two triangles |
| Source | Official CBSE paper or archive |
| Solution | Complete worked answer |
Also Check: Class 10 Maths Previous-Year Papers | CBSE Class 10 Maths Important Questions
The CBSE sample-paper archive contains Class 10 resources across multiple academic sessions and should be preferred over unsourced third-party “previous-year” lists.
A simple clinometer activity shows how angles and measured distances can be used to estimate an inaccessible height.
Carry out this activity only from a safe, level location away from traffic, building edges, unstable ground and other hazards. A teacher or responsible adult should supervise outdoor measurement.
Formula:
Height above eye level = horizontal distance × tan θ
Total height = height above eye level + observer height
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Class 10 applications of trigonometry use right-triangle ratios to calculate heights, horizontal distances and lines of sight. Common examples involve towers, buildings, trees, shadows and lighthouses.
An angle of elevation is measured upward from the observer’s horizontal line, while an angle of depression is measured downward from the horizontal. Both are measured from a horizontal line, not a vertical line.
Tangent is usually used when vertical height and horizontal distance are involved because tan θ = opposite/adjacent. Sine or cosine is used when the hypotenuse or line-of-sight distance is involved.
Identify the side you know and the side you need. Use sine for opposite and hypotenuse, cosine for adjacent and hypotenuse, and tangent for opposite and adjacent.
Mark the observer, draw the horizontal and vertical lines, add the line of sight, place the angle at the observer and label all known lengths. Mark the right angle where a vertical object meets level ground.
Many problems give a horizontal distance and ask for a vertical height, or give the height and ask for the horizontal distance. Tangent directly connects those opposite and adjacent sides.
Calculate the vertical height above the observer’s eye level first, then add the given eye height. Do this only when the right triangle begins at the observer’s eye rather than at ground level.
Draw the two right triangles, give each horizontal distance a correct expression and form one tangent equation for each angle. Solve the equations together to find the shared height or unknown distance.
The current curriculum specifies simple height-and-distance problems using 30°, 45° and 60°. Students should check the official syllabus for their examination session before relying on older notes.
The current CBSE curriculum states that heights-and-distances problems should involve no more than two right triangles. Problems with more triangles should be treated as extension material unless the syllabus changes.
Yes. It forms part of the Trigonometry unit, which has a stated weightage within the Class 10 Mathematics theory paper. The unit weightage includes ratios, identities and applications rather than this chapter alone.
Students should prepare to use exact standard values and perform the required calculations without depending on a calculator. Check the instructions printed on the official question paper and current CBSE examination rules.
Download the Some Applications of Trigonometry Class 10 notes from Infinity Learn website. It have the complete CBSE Class 10 Chapter Wise detailed notes with formula sheet, diagram worksheet, important questions and answer key.
Yes, the definitions, diagrams, standard ratios and core height-and-distance methods are useful for both. Students should also practise the difficulty and question formats shown in the official sample paper for their chosen course.
Common mistakes include placing an angle of depression against a vertical line, choosing the wrong ratio, forgetting observer height, confusing distance segments and rounding √3 too early.
Exercise 9.1 develops the core textbook method, but students should also practise diagrams, sample-paper formats, case-based questions and verified examination questions. Use official sample papers to understand current presentation and difficulty.
