Some Applications of Trigonometry Class 10 Notes (2026-27) — CBSE
Class 10 Maths Chapter 9 notes: heights and distances using angles of elevation and depression, line of sight, and the standard-angle method for solving problems.
Some Applications of Trigonometry — Class 10 Maths Notes
Chapter Snapshot
This chapter puts the trigonometric ratios of Chapter 8 to work in real-life "heights and distances" problems — finding the height of a tower, the width of a river, or the distance of a ship, using angles of elevation and depression. Every problem reduces to a right triangle you solve with sin, cos, or (most often) tan.
Board relevance: almost always gives one 3–4 mark application problem. A clear labelled figure earns marks and prevents mistakes.
Key Concepts & Definitions
Line of sight — the line drawn from the observer's eye to the object being viewed.
Horizontal line — the horizontal level through the observer's eye.
Angle of elevation — the angle between the line of sight and the horizontal when the object is above the horizontal (you look up).
Angle of depression — the angle between the line of sight and the horizontal when the object is below the horizontal (you look down).
Key fact: the angle of depression from a top point equals the angle of elevation from the bottom point, because the two horizontal lines are parallel (alternate interior angles).
Formulas
Every problem is a right triangle. With θ the angle and the right angle at the base:
To relate Use
Height (opposite) and base distance (adjacent) tan θ = height / distance
Height (opposite) and line of sight (hypotenuse) sin θ = height / hypotenuse
Base distance (adjacent) and line of sight (hypotenuse) cos θ = distance / hypotenuse
Standard-angle values you will use constantly:
θ tan θ sin θ cos θ
30° 1/√3 1/2 √3/2
45° 1 1/√2 1/√2
60° √3 √3/2 1/2
Method:
1. Draw a neat figure; mark the right angle, the given angle, and the known/unknown sides.
2. Choose the ratio that connects the known and unknown sides.
3. Substitute the standard value and solve. Rationalise the denominator if needed (√3 ≈ 1.732).
Worked Examples
Example 1 — Height of a tower: The angle of elevation of the top of a tower from a point 30 m away on the ground is 30°. Find the height.
tan 30° = h/30 → 1/√3 = h/30 → h = 30/√3 = 10√3 m ≈ 17.32 m.
Example 2 — Angle of depression: From the top of a 50 m cliff, the angle of depression of a boat is 45°. How far is the boat from the base?
The depression 45° equals the elevation at the boat. tan 45° = 50/d → 1 = 50/d → d = 50 m.
Example 3 — Two angles (find height): The angles of elevation of the top of a tower from two points at distances 4 m and 9 m from the base (on the same side) are complementary. Find the height.
Let the angles be θ and 90° − θ. tan θ = h/9 and tan(90° − θ) = cot θ = h/4.
Multiply: tan θ · cot θ = (h/9)(h/4) → 1 = h²/36 → h² = 36 → h = 6 m.
Example 4 — Two positions of observation: A man on a 20 m building sees a car's angle of depression change from 30° to 60° as it approaches. Find the distance the car travels.
At 30°: horizontal distance = 20/tan30° = 20√3. At 60°: = 20/tan60° = 20/√3.
Distance travelled = 20√3 − 20/√3 = (60 − 20)/√3 = 40/√3 = 40√3/3 m.
Example 5 — Width of a river: From a point on a bridge across a river, the angles of depression of the two opposite banks are 30° and 45°. If the bridge is 3 m above the banks, find the width of the river.
The two banks are on opposite sides, forming two right triangles sharing the height 3 m.
Bank A: tan 45° = 3/d₁ → d₁ = 3 m. Bank B: tan 30° = 3/d₂ → d₂ = 3√3 m.
Width = d₁ + d₂ = 3 + 3√3 = 3(1 + √3) ≈ 8.20 m.
This "two banks / opposite directions" figure — where the horizontal distances add rather than subtract — is a favourite twist, so always check whether the objects are on the same side or opposite sides of the observer.
A note on choosing the ratio: you never need all three ratios in one step. Look at which two sides the problem connects — the vertical height, the horizontal distance, or the slanting line of sight — and pick the single ratio that uses exactly those two. Using tangent whenever the hypotenuse is not involved keeps the arithmetic simplest, because you avoid square roots from the hypotenuse. Marking the known side and the unknown side on your figure before writing any equation almost always makes the correct ratio obvious.
Important Question Patterns
1. Single triangle (2–3 marks): height or distance from one angle of elevation/depression — pick tan/sin/cos and substitute the standard value.
2. Two angles from two points (3–4 marks): two right triangles sharing the height; write two tan equations and solve together (subtract distances or use complementary angles).
3. Angle of depression (3 marks): convert the depression angle to the equal elevation angle at the object, then solve.
4. Moving object (3–4 marks): distance moved between two observed angles = difference of the two horizontal distances.
5. Combined figure (4–5 marks): a pole on a building, or two towers — split into right triangles and add/subtract heights.
⚡ Quick Revision
- Angle of elevation = look up; angle of depression = look down (from horizontal). They are equal (alternate angles) between a top and bottom point.
- Right triangle: tan θ = height/distance (most common), sin θ = height/hypotenuse, cos θ = distance/hypotenuse.
- Standard angles: tan30 = 1/√3, tan45 = 1, tan60 = √3; sin/cos from the values table.
- Method: draw figure → mark right angle & given angle → choose the ratio linking known & unknown → substitute & solve; rationalise (√3 ≈ 1.732).
- Two-angle problems: two triangles, same height → solve the pair together. Complementary angles → tan θ · cot θ = 1.
- Distance moved = difference of horizontal distances at the two angles.
Get Started Free | Features | Pricing | Blog