Some Applications of Trigonometry Class 10: Notes and Questions

This guide covers Some Applications of Trigonometry Class 10, chapter 9 of NCERT Class 10 Mathematics, which is all about heights and distances. You will learn what the line of sight, angle of elevation and angle of depression are, and how to use tan to find heights of towers, widths of rivers and times for moving objects. Important questions with full solutions and common mistakes are at the end.

Line of sight and angle of elevation

The line of sight is the imaginary line from the observer's eye to the point being viewed.

The angle of elevation is the angle between the line of sight and the horizontal when the object is above eye level. It is the angle through which you raise your head, for example to see the top of a tower or a bird in the sky.

Every problem becomes a right-angled triangle:

• the object (tower, building) is the vertical side,
• the distance from the observer is the base,
• the line of sight is the hypotenuse.

Angle of depression

The angle of depression is the angle between the line of sight and the horizontal when the object is below eye level. It is the angle through which you lower your head, like a girl on a balcony looking down at a flower pot on the street.

The horizontal line at the observer's eye is parallel to the ground. So the angle of depression from the observer equals the angle of elevation of the observer as seen from the object. They are alternate interior angles between parallel lines.

Tip: always draw the horizontal line from the observer's eye first, then mark the angle of depression from it. A common error is to mark the angle from the vertical wall instead.

Using tan to find heights

When the hypotenuse is not involved, use the tangent ratio, which links height and distance:

tan θ = Opposite / Adjacent = height / distance

Values you will use here: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.

Example: from a point 15 m from the foot of a tower, the angle of elevation of the top is 60°. Then tan 60° = h/15, so √3 = h/15 and h = 15√3 m.

When the observer's height matters

If the observer's height is given, the triangle starts at eye level, not at the ground. Find the vertical side of the triangle, then add the observer's height to get the full height.

Example: an observer 1.5 m tall stands 28.5 m from a chimney. The angle of elevation of the top is 45°. In the triangle, tan 45° = AE/28.5. Since tan 45° = 1, AE = 28.5 m. Total height = 28.5 + 1.5 = 30 m.

Two triangles with one shared side

Many problems have two right triangles sharing the same vertical side. Write one equation for each angle, then substitute.

Shadow example: a tower's shadow is 40 m longer when the sun's altitude is 30° than when it is 60°. Let the height be h and the shorter shadow be x.

• At 60°: tan 60° = h/x, so h = x√3.
• At 30°: tan 30° = h/(x + 40), so h = (x + 40)/√3.
• Equate: x√3 = (x + 40)/√3. Multiply by √3: 3x = x + 40, so 2x = 40 and x = 20.
• h = 20√3 m.

For a fixed tower, the shadow length is h/tan θ. It is h√3 at 30° and h/√3 at 60°, so the shadow gets shorter as the sun rises higher.

Buildings and rivers

Two buildings: from the top of a taller building, the angles of depression to the top and to the bottom of a shorter building give two triangles. The difference in height between the two roofs is one part of the taller building's height.

River width: from a point on a bridge of known height, look at both banks. If the banks are on opposite sides of the observer, add the two horizontal distances. If they are on the same side, subtract the smaller from the larger.

Example: from a bridge 3 m high, the angles of depression of the opposite banks are 30° and 45°.

• Bank 1: tan 30° = 3/d₁, so d₁ = 3√3 m.
• Bank 2: tan 45° = 3/d₂, so d₂ = 3 m.
• Width = 3√3 + 3 = 3(√3 + 1) m.

Trigonometry with motion

When an object moves at uniform speed towards a tower, distance covered is proportional to time taken.

1. Draw the two triangles for the start and end positions.
2. Write both distances from the tower in terms of h using tan.
3. Find the distance covered in the given time.
4. Use Distance = Speed × Time to find the time for the rest of the way.

Remember this

• Elevation: look up. Depression: look down. Both are measured from the horizontal.
• Angle of depression from A to B = angle of elevation from B to A.
• Use tan θ = height/distance in most problems.
• Add the observer's height at the end when it is given.
• Two angles, one height: write two equations and substitute.

Important questions with answers

1. You look at a bird in the sky. Is the angle with the horizontal an angle of elevation or depression? Elevation, because the bird is above your eye level.

2. Why does the angle of depression from a cliff to a boat equal the angle of elevation from the boat to the cliff? The horizontal at the cliff top is parallel to sea level, so the two angles are alternate interior angles.

3. A 1.2 m tall girl sees a balloon. The triangle gives a height of 87 m above her eyes. How high is the balloon above the ground? 87 + 1.2 = 88.2 m.

4. An observer's eye is 100 m above sea level and sees a boat at an angle of depression of 30°. How far is the boat from the cliff base? tan 30° = 100/d, so 1/√3 = 100/d and d = 100√3 m, about 173.2 m.

5. A tower's shadow is 40 m longer at a sun altitude of 30° than at 60°. Find its height. As solved above: x = 20 m and h = 20√3 m.

6. From the top of a 60 m building, the angles of depression of the top and bottom of a shorter building are 30° and 60°. Find the distance between the buildings and the height of the shorter one. Bottom: tan 60° = 60/d, so d = 60/√3 = 20√3 m. Top: the drop from the taller roof to the shorter roof is v, with tan 30° = v/d, so v = 20√3 × (1/√3) = 20 m. The shorter building is 60 − 20 = 40 m tall, and the buildings are 20√3 m apart.

7. A car moves at uniform speed towards a tower. Its angle of depression from the top changes from 30° to 60° in 6 seconds. How much longer will it take to reach the tower? At 30° the distance is h√3; at 60° it is h/√3. Distance covered = h√3 − h/√3 = 2h/√3, taking 6 s. Remaining distance = h/√3, which is half of that, so it takes 3 s.

8. Both banks of a river are on the same side of an observer. How do you find the width? Subtract the smaller horizontal distance from the larger one.

Common mistakes to avoid

• Measuring the angle of depression from the vertical instead of the horizontal.
• Forgetting to add the observer's height at the end.
• Mixing up tan 30° and tan 60°.
• Adding river distances when both banks are on the same side.
• Starting calculations without a clear labelled diagram.

Draw a clean diagram for each problem before any calculation, and study this chapter with Joy to build confidence with heights and distances.