Class 10 Mathematics Notes

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Some Applications of Trigonometry Class 10 Maths Notes

The last chapter of the trigonometry unit puts the ratios to work on real objects: a tower, a pole, a lighthouse, a cliff. The measurement is always taken from a horizontal line at the observer's own level, so every problem becomes one right triangle or two, and every angle is 30°, 45° or 60°.

Class:10Subject:MathematicsUnit:VCovers:CBSE 2024-25

6 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

What is the difference between an angle of elevation and an angle of depression?

Both are measured from the horizontal line drawn through the observer's eye. The angle of elevation is measured upwards from that horizontal to the line of sight of the object above, while the angle of depression is measured downwards from it to the object below. And the angle of depression from A to B always equals the angle of elevation from B to A.

The Two Angles and How to Read Them

Definition

Angle of elevation

The angle between the horizontal line through the observer's eye and the line of sight of an object lying above that horizontal, measured upwards. It is the angle at the observer's eye inside the right triangle whose other vertex is the top of the object.

Definition

Angle of depression

The angle between the horizontal line through the observer's eye and the line of sight of an object lying below that horizontal, measured downwards. The horizontal line is taken at the level of the eye, which is not the ground, whenever the observer is standing on something.

Both angles are always measured from a horizontal, never from the ground and never from the object.
Elevation means the object is above the observer's horizontal; depression means it is below it.
The angle is at the observer's eye in both cases, so the observer's vertex is the correct vertex to look for in the figure.
The other two vertices of the right triangle are the foot of the perpendicular and the top or bottom of the object.
If the question says the angle of the sun, of a bird or of the top of a tower, it is the angle of elevation; if it says the angle of depression from a height, it is depression.

The rule that saves half the work

The angle of depression from A looking at B is always equal to the angle of elevation from B looking at A. So a problem giving a depression angle can be turned into an elevation problem before you touch a ratio.

Describing the Figure in Words

You will not always be given a diagram, so learn to build one from the words. Every problem in this chapter is the same picture with different objects: a vertical object standing on a horizontal surface, an observer somewhere on that surface, and one or two right triangles joining them.

The vertical object is the tower, pole, building, cliff or lighthouse. Call its foot B and its top A, so the height to be found is AB standing vertically.
The horizontal surface is the ground. Through the observer's eye draw a level line, and call its foot on the object D, so DB is also a height.
The ground distance from the observer to the foot of the object is the base of the right triangle; the line from the eye to the top of the object is the hypotenuse.
The angle at the observer's eye is the given angle, and it is an elevation when the top of the object is above the eye and a depression when it is below.
Two right triangles appear when the observer moves, or when the observer is at a height and sees both the top and the foot of the object; share the common horizontal distance between them.

Label before you calculate

Sketch the figure, mark the right angle at the foot of the perpendicular, write the given length and the given angle in their places, then label the unknown x. At that point the problem tells you which ratio to use, and the arithmetic is one line.

Why the Depression Angle Reverses

The reversal is not a coincidence and it is worth one line of justification, because a five-mark answer often needs it. Draw the figure: A is the top of the object, B its foot, C the observer's eye, and CD the horizontal line from the eye to the object, meeting it at D.

The angle of depression from C is ∠BCD, measured downwards from the horizontal CD to the line CB.
The angle of elevation from C to A is ∠DCA, measured upwards from the same horizontal CD to the line CA.
Since CD is horizontal and AB is vertical, AB and CD are both perpendicular to the same line, so AB ∥ CD.
With AB ∥ CD, ∠CBD and ∠BCD are alternate interior angles between the parallel lines and the transversal CB, so ∠BCD = ∠CBD.
And ∠CBD is the angle of elevation of the top of the object from B, the foot of the object. Hence depression from the top equals elevation from the foot.

Use it every time

Convert every depression angle in the problem into the corresponding elevation angle before you choose a ratio. A ship seen at 30° depression from a lighthouse top is the same 30° elevation of the lighthouse top from the ship, and the right triangle is unchanged in size.

Worked Problem 1 — A Single Right Triangle

Problem. From a point on a level ground, 12√3 m away from the foot of a tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower. Figure: the tower stands vertically with its foot at B on the ground; the observer stands at P on the same ground with BP = 12√3 m; PB is horizontal and the tower is vertical, so angle B is the right angle; P is the vertex of the given 60° angle, which is therefore an angle of elevation; AB is the height, call it h.

Solution

tan 6 0^{\circ} = \frac{h}{12 3} ⟹ h = 123 \times 3 = 12 \times 3 = 36 m

Both legs are known or wanted and the hypotenuse is not involved, so tangent is the ratio to use — opposite AB over adjacent PB.
Substitute tan 60° = √3 rather than 1.732, since no tables are supplied and the answer is meant to be exact.
Multiply straight through: 12√3 × √3 = 12 × 3 = 36, with no rationalising needed.
The single triangle is confirmed by the data: one length and one angle is exactly enough for a right triangle, which is why the problem is solvable at all.

Answer and check

The height of the tower is 36 m. Check: tan 60° = 36 / 12√3 = 36 / 20.78 = 1.732, and √3 = 1.732, so the ratio agrees.

The Method for a Two-Triangle Problem

A two-triangle problem shares one unknown height and gives you two different angles, or one angle and one distance change. The height is the same in both triangles, so the whole job is to write two equations in the height and eliminate one unknown. The syllabus caps a problem at two right triangles, so this is as complex as it ever gets.

Name the foot of the object B and its top A, then the two observer positions P and Q with P nearer to B. Let BP = x and let the distance PQ be the stated shift.
Draw the horizontal through the observer; both triangles are then right-angled at B, and BQ = BP + PQ = x + PQ.
Convert any depression angle into the matching elevation angle first, using the reversal rule.
Write the two equations from the same height: AB = BP × tan θ₁ and AB = BQ × tan θ₂.
Equate them, solve for x, and then substitute back into either equation for the height.
Keep surds exact throughout and rationalise only the final answer.

The classic slip in a two-triangle problem

Writing BQ as PQ instead of BP + PQ, or forgetting that the second observer is further away and therefore sees a smaller angle. Sanity check every two-triangle answer: the nearer observer must always see the larger angle.

Worked Problem 2 — Two Right Triangles

Problem. From a point P on a level ground the angle of elevation of the top of a tower is 60°. On moving a distance 20√3 m further away, to a point Q, the angle of elevation becomes 30°. Find the height of the tower. Figure: the tower is AB standing vertically on the ground with foot B and top A; P is the nearer observer and Q the further one, with Q beyond P; BP = x and PQ = 20√3 m, so BQ = x + 20√3 m; the angles at P and Q are both angles of elevation, 60° and 30°; let the height AB = h.

Two equations in the height h

tan 6 0^{\circ} = \frac{h}{x} ⟹ h = x 3 tan 3 0^{\circ} = \frac{h}{x + 20 3} ⟹ h = \frac{x + 20 3}{3} = \frac{x}{3} + 20

Equate, solve for x, then for h

x 3 = \frac{x}{3} + 20 ⟹ 3 x = x + 203 ⟹ x = 103, h = 103 \times 3 = 30 m

Both equations are for the same height h, so equating them removes h and leaves only x.
The second distance from the foot of the tower is x + 20√3, never 20√3 on its own.
Substituting x = 10√3 into the first equation gives h = 10√3 × √3 = 30, and substituting into the second gives h = 30√3 / √3 = 30 as well, so both triangles agree.

Answer

The height of the tower is 30 m, and the nearer observer stands 10√3 m from its foot while the further one stands 30√3 m away. The nearer observer sees 60° and the further one 30°, which is the order the sanity check demands.

A Second Two-Triangle Example — Two Ships

Problem. The angle of depression of two ships A and B from the top of a 60 m lighthouse, on the same side of it, are 60° and 30° respectively. Find the distance between the ships. Figure: the lighthouse is XY with X at the top and Y at the foot; the two ships lie on the same side along the sea level; draw the horizontal from X down to the sea level at Z, with Z beyond ship A; XZ = 60 m; the angles at X to the two ships are 60° and 30°, both angles of depression.

Distance to the nearer ship

tan 6 0^{\circ} = \frac{60}{Z A} ⟹ Z A = \frac{60}{3} = 203 m

Distance to the further ship and the gap

tan 3 0^{\circ} = \frac{60}{Z B} ⟹ Z B = 603 m, A B = Z B - Z A = 603 - 203 = 403 m \approx 69.3 m

Subtract, and check the order

Because the two ships lie on the same side, the distance between them is the difference ZB − ZA, not the sum. And the larger angle of depression, 60°, belongs to the nearer ship, so its distance is the smaller one, 20√3 m.

Common Errors in This Chapter

Almost every lost mark in heights and distances comes from one of a small number of habits. Read this list once before you start the paper.

Measuring from the ground instead of the horizontal through the observer's eye. When the observer is on a building or a cliff, this changes the whole triangle.
Using the angle of depression as if it were an angle of elevation without reversing it, and putting the observer at the wrong end of the triangle.
Choosing sine when tangent is needed. If you know both legs, or a leg and the other leg is wanted, tangent is the only ratio with the right shape — sine always brings in the hypotenuse.
Forgetting that the second distance is the sum of the first distance and the shift, so a two-triangle problem comes out with an impossible negative distance.
Subtracting the two distances instead of adding them when the two observation points lie on opposite sides of the object.
Quoting an answer in a form the question did not ask for, such as 1 / √3 where √3 / 3 is expected.
Attempting a problem with an angle other than 30°, 45° or 60°, which cannot be evaluated without a table that is not supplied in this paper.

Two habits that protect the marks

Label the figure before you write anything, and name the ratio in the answer. The mark for the ratio, the mark for the table value and the mark for the arithmetic are three separate marks, and an answer that jumps straight to the number collects only the last one.

How the Questions Are Asked

This chapter carries part of the 12 marks for the trigonometry unit and supplies the case-study questions that the 3-hour paper reserves for realistic situations. Each question is worth 4 marks in the case-study section, usually as two sub-parts.

Find the height of a tower when the angle of elevation and the distance from the foot are given. (A three-mark short answer.)
Find the distance from the foot of a tower when the height and the angle of elevation are given. (Three marks.)
A two-triangle problem with the observer moving, or seeing the top and the foot from a height. (A five-mark long answer.)
Explain why the angle of depression from A to B equals the angle of elevation from B to A. (Two or three marks.)
A case study on a lighthouse, a cliff, a pole or a cable-stayed bridge, with one sub-part on finding a height and one on finding a distance.

The syllabus cap on difficulty

A problem will not involve more than two right triangles, and the angles of elevation and depression used will be only 30°, 45° and 60°. If a problem seems to need a third triangle or a table, you have misread it rather than found a hard question.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Angle of elevation, height from base

tan θ = \frac{h}{d} ⟹ h = d tan θ

θ is the angle of elevation at the observer, d the ground distance from the foot of the object and h its height.

Ground distance from height and angle

d = \frac{h}{tan θ} = h cot θ

Use this when the height of the object is given and the distance must be found.

Depression equals the reverse elevation

∠ depression from A to B = ∠ elevation from B to A

The two horizontal lines are parallel, so the angles are alternate interior angles. Convert every depression angle before choosing a ratio.

Single triangle, one leg and the angle

sin θ = \frac{h}{l}, cos θ = \frac{d}{l}, tan θ = \frac{h}{d}

h is the vertical height, d the horizontal distance and l the sloping line of sight, which is the hypotenuse. A height or a ground distance needs only the two legs, which is why tangent is used so often.

Two-triangle elimination

x tan θ_{1} = (x + s) tan θ_{2}

With s the shift between the two observer positions, equating the two expressions for the height and solving for x. Then h = x tan θ₁.

The values used in every problem

tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3

Only these three angles are set, because no trigonometric tables are supplied in the paper.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Both angles are measured from the horizontal line through the observer's eye, not from the ground and not from the object.
The angle of depression from A to B always equals the angle of elevation from B to A, so convert every depression angle before choosing a ratio.
Use tangent when you know or want both legs, sine when the hypotenuse or the sloping line is involved, and cosine when you have the hypotenuse and want the horizontal leg.
In a two-triangle problem the second ground distance is the first plus the shift, and the nearer observer always sees the larger angle — use that as a check.
Only 30°, 45° and 60° are set, and no trigonometric tables are supplied, so keep surds exact and rationalise the final answer to √3 / 3 rather than 1 / √3.
A problem will never need more than two right triangles; if yours seems to need a third, you have misread the data.
Name the ratio and state the table value before substituting, because the mark for the ratio and the mark for the arithmetic are separate.
A two-mark question often only asks why the depression and the reverse elevation are equal, and that answer is the alternate-angles argument on parallel lines.
Case-study questions in Section E carry 4 marks each as two sub-parts, and they use the same method as a textbook question with a story attached.
Trigonometry carries 12 of the 80 theory marks, and internal assessment is 20 marks — 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical — with no chapter-wise split.

FAQ

Frequently asked questions

What is the difference between an angle of elevation and an angle of depression?

Both are measured from the horizontal line drawn through the observer's eye. The angle of elevation is measured upwards from that horizontal to an object above the observer's level, and the angle of depression is measured downwards from it to an object below that level. The two angles are complementary to nothing in particular, but the depression from A to B is always equal to the elevation from B to A.

Why is the angle of depression from A to B equal to the angle of elevation from B to A?

Draw the horizontal through the observer's eye so that it meets the vertical object, and call the vertical and the horizontal two parallel lines, since both are perpendicular to the ground. The line of sight acts as a transversal cutting them, so the angle of depression at the observer and the angle of elevation at the foot of the object are alternate interior angles and are therefore equal.

A tower is observed from a point 12√3 m away from its foot at an angle of elevation of 60°. Find its height.

The horizontal distance from the foot of the tower is 12√3 m and the height is the side opposite the 60° angle, while the known distance is the side adjacent to it, so tangent is the ratio to use. Hence tan 60° = h / 12√3, giving h = 12√3 × √3 = 36 m. The answer is exact because 60° is one of the three standard angles.

How do I solve a problem with two right triangles?

Name the foot of the object B, the two observer positions P and Q with P nearer, and let BP = x with PQ = s, so BQ = x + s. Convert any depression angle into the matching elevation angle. Write the height from each triangle, h = x tan θ₁ and h = (x + s) tan θ₂, equate them to remove h, and solve for x. Substitute back for h. For a tower seen at 60° from one point and 30° from a point 20√3 m further away, this gives x = 10√3 m and a height of 30 m.

What mistakes are most common in heights and distances questions?

The most frequent are: measuring the angle from the ground instead of the horizontal through the eye; failing to reverse an angle of depression; using sine when tangent is required; taking the second ground distance as the shift alone instead of the first distance plus the shift; subtracting two distances that should be added when the observation points are on opposite sides of the object; and using an angle other than 30°, 45° or 60°, which cannot be evaluated without a table that is not supplied in the paper.

Next Chapters

Continue learning

Class 10 Maths Notes Introduction to Trigonometry Class 10 Maths Notes Trigonometric Identities Class 10 Maths Notes Triangles Class 10 Maths Notes All Class 11 & 12 Notes

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