Some Applications of Trigonometry – Class 10 Maths Chapter 9 Notes & NCERT PDF

Class 10MathematicsChapter 9NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 9 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 9: Some Applications of Trigonometry

Introduction

In this chapter we use trigonometric ratios to find heights and distances that cannot be measured directly — the height of a tower, the width of a river, the length of a ladder. Surveyors have used such methods since ancient times.

Key Concepts

Line of sight

The line drawn from the eye of an observer to the point on the object being viewed.

Angle of elevation

When the object is above the horizontal level of the eye, the angle between the line of sight and the horizontal is the angle of elevation. The observer has to raise the head to see it.

Angle of depression

When the object is below the horizontal level of the eye, the angle between the horizontal and the line of sight is the angle of depression. The observer has to lower the head.

Because the horizontal through the eye is parallel to the ground, the angle of depression from A to B equals the angle of elevation from B to A (alternate angles).

What we need to solve a problem

The height or length of an object, or the distance between two distant objects, can be found using a right triangle when we know:

one side, and

one acute angle (the angle of elevation or depression).

Methods Step by Step

Draw a neat, labelled figure: a vertical line for the tower/pole/building, a horizontal line for the ground, and the line of sight.

Mark the right angle and the given angle(s).

Choose the ratio that links the known side and the unknown side — usually tan (opposite/adjacent) for height-and-distance, sin or cos when a slanting length (ladder, string, kite thread) is involved.

If the observer has a height, add or subtract it (e.g. observer's eye 1.5 m above the ground).

With two angles, write two equations and eliminate the common unknown.

Use √3 = 1.732 (or 1.73) only at the final step if asked.

Worked Examples (from the book)

Tower, 15 m away, elevation 60°. tan 60° = AB/15 ⇒ AB = 15√3 m.

Electrician's ladder. Pole 5 m; repair point 1.3 m below the top, so BD = 3.7 m. Ladder at 60°: sin 60° = 3.7/BC ⇒ BC = 7.4/√3 ≈ 4.28 m. Distance of foot from pole: tan 60° = 3.7/DC ⇒ DC = 3.7/√3 ≈ 2.14 m (using √3 = 1.73).

Observer 1.5 m tall, 28.5 m from a chimney, elevation 45°. tan 45° = AE/28.5 ⇒ AE = 28.5 m; height = 28.5 + 1.5 = 30 m.

10 m building, elevation 30°; top of flagstaff at 45°. tan 30° = 10/PA ⇒ PA = 10√3 ≈ 17.32 m. tan 45° = (10 + DB)/PA ⇒ DB = 10√3 – 10 = 10(√3 – 1) ≈ 7.32 m.

Shadow is 40 m longer when the Sun's altitude is 30° than at 60°. With height h and shorter shadow x: h = x√3 and h/(x + 40) = 1/√3 ⇒ x = 20, h = 20√3 m.

Angles of depression 30° and 45° of the top and bottom of an 8 m building. Height of the multi-storeyed building = 4(3 + √3) m, and the distance between the buildings is also 4(3 + √3) m.

Bridge 3 m high; angles of depression of the two banks 30° and 45°. AD = 3√3 m and DB = 3 m ⇒ width of river = 3(√3 + 1) m.

Board Exam Focus

Almost every board paper has one heights-and-distances question (3–5 marks). Marks are given for the figure, the choice of ratio, and the final answer with units.

Two-angle problems (shadow, flagstaff, two buildings, river width) are the most common long-answer type.

Always state which triangle you are using.

Formula Sheet

tan θ = height / horizontal distance

sin θ = height / slant length (ladder, string)

cos θ = horizontal distance / slant length

tan 30° = 1/√3, tan 45° = 1, tan 60° = √3

sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2

√3 ≈ 1.732, √2 ≈ 1.414

Common Mistakes

Measuring the angle of depression from the vertical instead of from the horizontal.

Forgetting to add the observer's own height.

Using sin when neither the given nor the required side is the hypotenuse.

Putting the angle at the wrong vertex in the figure.

Rounding √3 too early, giving inaccurate final answers.

Writing answers without units (m).

💡 Key Learning Points

✓Define line of sight, angle of elevation and angle of depression, and tell them apart in real situations.
✓Recognise that the angle of depression from one point equals the angle of elevation from the other (alternate angles).
✓Draw a correct labelled figure and choose the right ratio (usually tan; sin/cos for slanting lengths).
✓Solve single-angle and two-angle heights-and-distances problems, including observer height.
✓Present answers with correct units and use √3 ≈ 1.732 only at the end.

👨‍🏫 Teaching Tips

→Take students outside with a simple clinometer (protractor + straw + thread) to measure the angle of elevation of the school building and compute its height.
→Insist that every solution starts with a neat figure; award separate marks for the figure in class tests.
→Use a paper strip model to show that the angle of depression and angle of elevation are alternate angles.
→Solve the shadow problem (30° and 60°) with algebra on the board, showing how to eliminate the common unknown.
→Give a 'which ratio?' drill: students only name the ratio needed for 10 quick situations.
→Relate to careers — surveyors, architects, navy and air-traffic controllers use these methods.

📋 Assessment Questions

Can the student draw a correct figure for a two-angle problem?
Can the student identify whether the given angle is elevation or depression?
Can the student solve the flagstaff-on-building problem to get 10(√3 – 1) m?
Does the student remember to add observer height where needed?
Are final answers given with units and correct approximation?

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