Some Applications of Trigonometry Class 10 Notes: Chapter 9

These Notes for Class 10 Maths Chapter 9 Some Applications of Trigonometry give you a quick, concept-first revision of the whole chapter. It uses the trigonometric ratios from Chapter 8 to find heights and distances you cannot measure directly: how tall a tower is, how far a ship has sailed, or how high a kite is flying.

• Angle of elevation and angle of depression made clear, with diagrams, solved examples, and a step-by-step method for every board problem type.
• Full coverage of how to turn a word problem into a right-angled triangle, pick the correct ratio, and solve for the unknown.
• Aligned to the 2026-27 CBSE syllabus and written to help you score full marks in the heights and distances section.

These Collegedunia revision notes are curated by Maths subject experts, according to the 2026-27 NCERT textbook, and refined against the last five years of CBSE Class 10 Maths board papers.

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What These Notes Cover

Chapter 8 taught the trigonometric ratios; Chapter 9 uses them to find heights and distances you cannot measure directly. A lighthouse, a kite, a flagpole, a river's width: all can be measured from afar using a known base and an observed angle.

• Angle of elevation: the angle you look up through, from the horizontal to an object above.
• Angle of depression: the angle you look down through to an object below (say, from a cliff).
• Single-triangle problems: one observer, one object. Draw the triangle, mark the sides, apply the right ratio.
• Two-angle problems: two angles from one or two positions. Form two equations and solve together.

The whole chapter rests on one skill: turning a word description into a right-angled triangle, then reading off the right ratio.

Angle of Elevation and Angle of Depression

These two terms set up almost every problem in the chapter.

Angle of elevation

Stand at O and look up at an object P. The angle of elevation is the angle θ between the horizontal at O and the line of sight OP. In the right-angled triangle, the height is the opposite side, the distance is the adjacent side, and the line of sight is the hypotenuse. So tan(angle of elevation) = height / horizontal distance, which fits most problems because height and distance are the usual givens.

Angle of depression

Now stand on a cliff at P and look down at an object O below. The angle of depression is the angle between the horizontal at P and the line of sight PO. Key fact: the angle of depression from P to O equals the angle of elevation from O to P (alternate interior angles), so you can swap between them and use the same triangle and ratio.

Setting Up a Heights and Distances Problem

Every problem follows the same five steps. Practise them until they are automatic.

tan θ = opposite / adjacent is used most, because height and distance are exactly the opposite and adjacent sides. When in doubt, use tan first.

Key Trigonometric Ratios

This chapter uses the standard values from Chapter 8. Board problems mostly use 30°, 45°, and 60°, which give clean surds.

At 45°, tan 45° = 1, so height = distance, the fastest sum with no surds. At 60°, height = d√3. At 30°, height = d/√3 = d√3/3 after rationalising.

Solved Examples: Heights and Distances

These cover the main board types: single-angle elevation, depression from a height, and the two-angle walking problem. Try each with the five-step setup first.

Example 1: Height of a tower (single angle)

Problem: From a point 50 m from a tower's base, the angle of elevation of the top is 60°. Find the height.

Solution: Let the height be h. tan 60° = h / 50, so √3 = h / 50, giving h = 50√3 m.

Answer: The height of the tower is 50√3 m (about 86.6 m).

Example 2: Distance from the base (angle of depression)

Problem: From the top of a 75 m lighthouse, the angle of depression of a ship is 30°. How far is the ship from the base?

Solution: By alternate angles, the elevation from the ship is also 30°. With d the horizontal distance, tan 30° = 75 / d, so 1/√3 = 75 / d, giving d = 75√3 m.

Answer: The ship is 75√3 m (about 129.9 m) from the base.

Example 3: Two-angle problem (observer walks closer)

Problem: From a point P, the angle of elevation of a building's top is 30°. After walking 20 m toward it, the angle becomes 60°. Find the height.

Solution: Let the height be h and the distance to the nearer point Q be x, so P is (x + 20) away.

From P: tan 30° = h / (x + 20), so h = (x + 20)/√3 ... (i). From Q: tan 60° = h / x, so h = x√3 ... (ii).

Equating: x√3 = (x + 20)/√3, so 3x = x + 20, giving x = 10 m, and h = 10√3 m.

Answer: The height of the building is 10√3 m (about 17.3 m).

Two-Observer and Two-Object Problems

Board papers often add a harder variant: two observers look at one object from opposite sides, or one observer looks at two objects. These make two right-angled triangles sharing a side (usually the tower height) that links the two equations.

• Two observers, same side: like Example 3. Same height h, different distances; equate the two expressions for h.
• Two observers, opposite sides: both look up; the two distances add up to the total between them.
• One observer, two objects: two depression angles, two triangles.

The opposite-side setup is the most common 4-mark question here: label the shared height h, label the two distances, write two tan equations, and solve, never combining the triangles into one equation.

How to Revise for the Board Exam

This is a skills chapter: the idea is narrow, but it needs steady practice. A three-pass plan works well:

• First, master the diagram. Take five problems from Exercise 9.1 and draw the diagram only. Check the right angle is at the foot of the vertical, the angle is at the observer, and the sides are labelled. A correct diagram makes the solution almost automatic.
• Then run the five-step method. Write the equation, substitute the standard value, and solve in two or three clean lines. Examiners reward clear working, so do not skip steps.
• Finally, drill two-angle questions. These carry more marks; do at least three timed, no notes, checking each before the next.

Previous Year Question Trends

This chapter is a reliable source of 3-mark and 4-mark questions, and the types repeat across years.

Also Check: The full set of board questions for this chapter, with step-by-step solutions, is in the Notes PDF above, updated for 2026-27.

Common Mistakes to Avoid

Most lost marks come from a short list of repeated errors. Knowing them before the exam is the fastest way to protect your score.

• No diagram or a wrong diagram: the right angle goes at the foot of the vertical, not the observer or top. Wrong placement means a wrong equation.
• Using sin or cos instead of tan: most problems give only height and distance, which tan links directly; sin and cos need the rarely-given hypotenuse.
• Not rationalising surds: leaving 75/√3 instead of 25√3 loses a mark.
• One variable for both distances: if the observer walks 20 m closer, use d and (d + 20), not d twice.
• Forgetting units: state every height and distance in metres.

Other Resources for This Chapter

Pair these notes with the matching NCERT Solutions, formula sheet, handwritten notes, and the official NCERT book chapter. All resources for Some Applications of Trigonometry are linked below.

Notes for Class 10 Maths: All Chapters

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