Class 10 — Mathematics (NCERT)

Some Applications of Trigonometry

Class 10

✓To understand the concepts of line of sight, angle of elevation, and angle of depression.
✓To apply trigonometric ratios to solve problems involving heights and distances of various objects.
✓To develop skills in drawing appropriate diagrams for real-world problems.
✓To solve practical problems related to heights and distances using trigonometric identities and ratios.

Key concepts

Line of Sight

The line of sight is the line drawn from the eye of an observer to the object being viewed.

Angle of Elevation

When an observer looks at an object that is above the horizontal level, the angle formed by the line of sight with the horizontal is called the angle of elevation.

Angle of Depression

When an observer looks at an object that is below the horizontal level, the angle formed by the line of sight with the horizontal is called the angle of depression.

Heights and Distances

This topic deals with finding the heights of various objects (like towers, buildings, trees) or distances between objects (like width of a river, distance of a ship from a lighthouse) using the knowledge of trigonometric ratios and angles of elevation or depression. It primarily involves solving right-angled triangles.

Key facts to remember

1The line of sight is the line from the observer's eye to the object.
2The angle of elevation is formed when the object is above the horizontal level of the observer's eye.
3The angle of depression is formed when the object is below the horizontal level of the observer's eye.
4In problems involving angles of depression, the angle of depression is equal to the angle of elevation from the object's position to the observer's eye (alternate interior angles).
5Trigonometric ratios (sin, cos, tan) are used to relate the sides and angles of a right-angled triangle.
6Always draw a neat, labelled diagram to represent the problem statement.
7Recall the values of trigonometric ratios for standard angles: 0°, 30°, 45°, 60°, 90°.

Worked examples

Example 1

A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

ILet AB be the height of the tower and C be the point on the ground.

IIGiven, BC = 15 m and ∠ACB = 60°.

IIIIn the right-angled triangle ABC, we have tan C = AB/BC.

IVSubstitute the values: tan 60° = AB/15.

VWe know tan 60° = √3. So, √3 = AB/15.

VIMultiply both sides by 15: AB = 15√3 m.

Answer

The height of the tower is 15√3 m.

Example 2

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

ILet A be the point on the bridge and BC be the width of the river. Let AD be the height of the bridge, so AD = 3 m.

IIThe angles of depression are ∠XAB = 30° and ∠YAC = 45°. (Draw a horizontal line XY through A).

IIISince XY is parallel to BC, ∠ABD = ∠XAB = 30° (alternate interior angles) and ∠ACD = ∠YAC = 45° (alternate interior angles).

IVIn right-angled ΔADB, we use tan 30° = AD/BD.

VSubstitute values: 1/√3 = 3/BD. Hence, BD = 3√3 m.

VIIn right-angled ΔADC, we use tan 45° = AD/CD.

VIISubstitute values: 1 = 3/CD. Hence, CD = 3 m.

VIIIThe width of the river BC = BD + CD.

9BC = 3√3 + 3 = 3(√3 + 1) m.

Answer

The width of the river is 3(√3 + 1) m.

It is crucial to draw a clear diagram and correctly identify the alternate interior angles for angles of depression.

Example 3

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

ILet AB be the building, so AB = 30 m. Let PQ be the boy's initial position, and RS be his final position.

IIThe boy's height is 1.5 m. Draw a horizontal line PE from the boy's eye level (P and R are eye positions).

IIIThe effective height of the building for calculation is AE = AB - EB = AB - PQ = 30 - 1.5 = 28.5 m.

IVLet the initial distance from the building be QC = x and the distance walked be QR = y. So, RC = x - y.

VIn right-angled ΔAEP, ∠APE = 30°. We use tan 30° = AE/PE.

VISubstitute values: 1/√3 = 28.5 / PE. So, PE = 28.5√3 m. (Note that PE = QC = x).

VIIIn right-angled ΔAER, ∠ARE = 60°. We use tan 60° = AE/RE.

VIIISubstitute values: √3 = 28.5 / RE. So, RE = 28.5/√3 = (28.5√3)/3 = 9.5√3 m. (Note that RE = RC = x - y).

9The distance walked is y = PE - RE.

10y = 28.5√3 - 9.5√3 = (28.5 - 9.5)√3 = 19√3 m.

Answer

The distance the boy walked towards the building is 19√3 m.

Remember to subtract the observer's height from the total height of the object if the angle is measured from eye level.

Common mistakes

✗Confusing the angle of elevation with the angle of depression, or placing them incorrectly in the diagram.
✗Not drawing a clear and accurate diagram, leading to incorrect identification of sides and angles.
✗Using the wrong trigonometric ratio (e.g., using sin instead of tan) for the given sides and angle.
✗Making calculation errors, especially with square roots (e.g., √3, √2).
✗Forgetting to adjust the height of the object when the observer's height is given and the angle is measured from the eye level.

Exam tips

★Read the problem carefully and draw a neat, labelled diagram representing the situation. This is the most crucial step.
★Identify the right-angled triangles in your diagram and label the known and unknown sides and angles.
★Choose the appropriate trigonometric ratio (sin, cos, or tan) that relates the known values to the unknown value you need to find.
★Substitute the values of trigonometric ratios for standard angles (30°, 45°, 60°) accurately.
★Show all steps clearly, including the formula used and the final answer with correct units. Hence proved.

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