Chapter 9 of 14

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CBSE Class 10 Mathematics★ Unit V — Trigonometry: 12 marks, shared with Introduction to Trigonometry · CBSE Class 10 Maths (Chapter 9)

Some Applications of Trigonometry

Some Applications of Trigonometry — heights and distances. Angle of elevation, angle of depression. Solve real-world problems involving towers, buildings, and aerial measurements. CBSE Class 10 Maths Chapter 9.

⏱ 20 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Draw a labelled right triangle from a worded situation
2Tell an angle of elevation from an angle of depression, and mark each correctly
3Choose the ratio that connects what you know to what you want
4Handle two-triangle problems where a common side links the two equations
5Account for the observer's own height when the question gives it

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Why this chapter matters

This is the chapter where trigonometry stops being algebra and starts measuring the world. Every question is a real situation — a tower, a kite string, a ship at sea — reduced to a right triangle you can solve.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Introduction to Trigonometry

"Trigonometry's true power is measuring what we cannot directly reach — towers, mountains, stars."

1. About the Chapter

This chapter APPLIES trigonometric ratios (Chapter 8) to REAL-WORLD problems of heights and distances.

Key Concepts

Angle of Elevation
Angle of Depression
Line of sight, horizontal line
Solving problems systematically

2. Key Terms

Line of Sight

The line drawn from the OBSERVER's eye to the OBJECT being seen.

Horizontal Line

The horizontal level passing through the observer's eye.

Angle of Elevation

The angle between LINE OF SIGHT (going UP) and HORIZONTAL LINE.

Used when object is ABOVE observer
Example: looking up at a tower

Angle of Depression

The angle between LINE OF SIGHT (going DOWN) and HORIZONTAL LINE.

Used when object is BELOW observer
Example: looking down from a balcony

Important Note

Angle of elevation = Angle of depression (alternate interior angles when looking from one to the other).

3. Solving Height-Distance Problems

General Procedure

DRAW DIAGRAM: sketch the situation with observer, object, angle
IDENTIFY known and unknown quantities
SET UP trigonometric ratio (usually tan, sometimes sin/cos)
SOLVE for unknown
VERIFY answer

Standard Setup

For a tower of height h, observer at distance d, angle θ: tan θ = h / d

So:

If d and θ known: h = d × tan θ
If h and θ known: d = h / tan θ
If h and d known: tan θ = h/d → θ = arctan(h/d)

4. Worked Examples

Example 1: Find Height

A man 1.8 m tall stands 30 m from a tower. Angle of elevation of top of tower is 60°. Find tower height.

Setup:

Man's height = 1.8 m
Horizontal distance = 30 m
Angle of elevation = 60°

Solve:

Height of tower above man's eye = 30 × tan 60° = 30√3 m
Total tower height = 30√3 + 1.8 ≈ 30(1.732) + 1.8 = 51.96 + 1.8 = 53.76 m

(For Class 10, often ignore observer's height for simplicity.)

Example 2: Find Distance

A boy looks at the top of a 30 m tall building at an angle of elevation of 30°. Find the distance from the building.

tan 30° = 30 / d
1/√3 = 30 / d
d = 30√3 ≈ 52 m

Example 3: Angle of Depression

From the top of a 100 m tall tower, a car on the ground is observed at an angle of depression of 45°. Find the distance of the car from the foot of the tower.

tan 45° = 100 / d
1 = 100 / d
d = 100 m

Example 4: Two Observation Points

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 100 m toward the tower, the angle becomes 60°. Find tower height.

Let height = h, initial distance = d.

tan 30° = h/d → h = d/√3
After walking 100m toward: tan 60° = h/(d−100) → h = √3(d−100)

Set equal: d/√3 = √3(d−100) d = 3(d−100) d = 3d − 300 −2d = −300 d = 150 m

h = 150/√3 = 50√3 ≈ 86.6 m

Example 5: Cliff and Boat

From a cliff 100 m high, the angles of depression of two boats are 30° and 45°. Find the distance between them.

Let distances from cliff base be d₁ (closer) and d₂ (farther).

tan 45° = 100/d₁ → d₁ = 100 m
tan 30° = 100/d₂ → d₂ = 100√3 ≈ 173 m

Distance between boats = d₂ − d₁ = 173 − 100 = 73 m (approx)

5. Tips for Word Problems

Always Draw a Diagram

Mark observer (eye level)
Mark object (with height labelled)
Mark angle clearly (elevation or depression)
Mark known/unknown distances

Identify Sides

For angle θ in right triangle:

Opposite (perpendicular)
Adjacent (base)
Hypotenuse

Choose Right Ratio

tan = opposite/adjacent (most common for heights/distances)
sin = opposite/hypotenuse (when hypotenuse mentioned)
cos = adjacent/hypotenuse (when hypotenuse mentioned)

Common Pitfalls

Don't confuse 'angle of elevation' (up) and 'angle of depression' (down)
Distance is usually HORIZONTAL distance (along ground)
Be careful with units

6. Real-World Applications

Civil Engineering

Building heights from ground
Bridge angles
Slope of roads

Surveying

Land surveys use theodolite (measures angles)
Indian Survey of India uses trigonometric methods

Aviation

Plane altitude estimation
Distance from runway

Astronomy

Star altitudes
Sun's angular elevation throughout day

Indian Context

Trigonometric Survey of India (1802-1871) mapped India
Famous: George Everest (mountain named after him)
Modern: GPS uses trigonometry

7. Worked Example with Multiple Heights

Example: Pole and Tower

A pole 6 m high casts a shadow 8 m long. At the same time, a tower casts a shadow 24 m long. Find the tower's height.

Using similar triangles (sun's angle same):

Pole: angle θ such that tan θ = 6/8 = 3/4
Tower: tan θ = h/24
So h/24 = 3/4
h = 18 m

(Same angle, so ratios are equal.)

8. Common Mistakes

Elevation vs Depression confusion
- LOOKING UP at object → elevation
- LOOKING DOWN at object → depression
Wrong trigonometric ratio
- Use TAN for most height-distance problems (no hypotenuse).
Ignoring observer height
- In some problems, observer's height matters. Read carefully.
Direction of walking
- 'Walking toward' decreases distance; 'walking away' increases.
Final answer should be POSITIVE
- If you get negative distance, recheck setup.

9. Conclusion

Trigonometry's REAL POWER is in measuring what we cannot directly access:

Tower and building heights
Mountain altitudes
Distances across rivers
Aircraft elevation
Sun and star positions

Master:

Drawing clear diagrams
Identifying angles correctly (elevation/depression)
Choosing right ratios (usually tan)
Solving with specific angle values

Practice 15+ problems to gain fluency.

Trigonometry: turning the unmeasurable into measurable.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Angle of elevation

angle between the line of sight and the horizontal, looking up

Measured at the observer, from the horizontal — not from the vertical

Angle of depression

angle between the line of sight and the horizontal, looking down

Equal to the angle of elevation from the object back to the observer (alternate angles)

Height from distance

height = distance × tan θ

Use when the horizontal distance is known

Distance from height

distance = height ÷ tan θ

Use when the vertical height is known

Slant length

hypotenuse = height ÷ sin θ

For ropes, strings, ladders and slides — the sloping line itself

The 30-60 pair

tan 30° = 1/√3 and tan 60° = √3

Almost every question in this exercise uses 30°, 45° or 60°

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Measuring the angle from the vertical

✓ Both elevation and depression are measured from the *horizontal*. Drawing the horizontal dashed line at the observer's eye first prevents this.

WATCH OUT

✗ Marking the angle of depression inside the triangle at the wrong vertex

✓ The depression angle sits between the horizontal at the top and the line of sight going down. Its equal partner inside the triangle is the angle of elevation at the object — use that one in your ratios.

WATCH OUT

✗ Forgetting the observer's height

✓ When a question says the boy is 1.5 m tall or the girl is 1.2 m tall, subtract it from the object's height first. The triangle sits on the level of the eyes, not the ground.

WATCH OUT

✗ Using sin where tan is needed

✓ tan links the two legs (height and horizontal distance); sin links a leg to the hypotenuse. Ropes and strings are hypotenuses, so those questions want sin.

WATCH OUT

✗ Leaving the answer as a decimal when a surd was expected

✓ Answers such as 8√3, 20(√3 − 1) and 75(√3 − 1) are exact. Give the surd, then the decimal if you wish — not the other way round.

WATCH OUT

✗ Solving two-triangle problems one triangle at a time

✓ The two triangles share a side. Write an expression for that shared side from each, set them equal, and solve — that single equation is the whole method.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 9.1

Exercise 9.1

Heights and distances — single-triangle problems, two-triangle problems, and observers with their own height

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Some Applications of Trigonometry?

3 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

3 questions~2 min worth ~10 marks in CBSE exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Line of sight, angle of elevation, angle of depression — the three definitions the summary names
•Angles of elevation and depression are always measured from the horizontal
•The angle of depression from A to B equals the angle of elevation from B to A
•tan for two legs, sin for a leg and the hypotenuse, cos rarely
•Two-triangle problems: find the shared side twice and equate
•Subtract the observer's height before you start, and note that the answer is then a height above eye level
•This chapter has a single exercise, 9.1, with 15 questions

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-10 marks

Question type	Marks each	Typical count	What it tests
MCQ	1	2	Definitions
Short	3	1	Single observation
Long	5	1	Two observation points

Prep strategy

Always draw diagrams
Practice 15+ problems
Master tan = h/d setup

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Survey of India

Used trigonometry to map India 1802-1871. Mountain Everest named after George Everest.

Engineering

Bridge angles, tower heights, slopes.

GPS

Modern positioning uses trigonometry.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Draw the figure first, every time — mark the horizontal, then the angle, then label the sides

Write down which side you know and which you want; that pair names the ratio

For two-triangle problems, name the shared side with a single letter and build both equations around it

Keep surds exact through the working and rationalise only at the end

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

3D trigonometry

STRETCH

Spherical trigonometry (Class 11)

STRETCH

Vector geometry

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 10 BoardVery High

Maths OlympiadHigh

JEEVery High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Both are measured from the horizontal at the observer. If you raise your head to look at the object, the angle between your line of sight and the horizontal is the angle of *elevation*; if you lower your head, it is the angle of *depression*. Usefully, the angle of depression from the top of a tower down to a car equals the angle of elevation from the car back up to the top, because they are alternate angles between parallel horizontals — which is why depression problems can always be redrawn as elevation problems.

Because those are the only angles whose ratios you know exactly from Chapter 8. Real surveying uses tables or a calculator, but at this level the questions are constructed so that every answer comes out as an exact surd.

Subtract it from the object's height and work with the difference. In Exercise 9.1 Q6 the boy is 1.5 m tall and the building 30 m, so the triangle has height 28.5 m; in Q14 the girl is 1.2 m and the balloon at 88.2 m, giving 87 m. The horizontal distances are unaffected, which is why the answers to those questions do not need the height added back.

The chapter kept its single exercise, but it now has 15 questions rather than the 16 of the older edition, and the separate Introduction section was folded in — the rationalised chapter is just 9.1 Heights and Distances followed by 9.2 Summary.

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