Chapter – 9

Some Applications of Trigonometry

In this post we have given the detailed notes of class 10 Maths Chapter 9 (Some Applications of Trigonometry) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 9: Some Applications of Trigonometry

Heights and Distances

Trigonometry has wide applications in solving problems related to heights and distances in real life. It is one of the most effective methods used for the indirect measurement of heights and distances that cannot be measured directly. Trigonometry is used extensively in geography, navigation, astronomy, and map-making.

Uses of Trigonometry

• Finding the height of a tall building or tower.
• Finding the width of a river or a sea.
• Finding the distance of planets and stars from the earth.
• Making maps and locating the position of an island in terms of its latitude and longitude.
• Finding the distance or height of a flying object from a given point.
• Used extensively in engineering and physics.

Important Terms

• Line of Sight — the line joining the eye of an observer to the point on the object being viewed.
• Horizontal Line — the line joining the foot of the observer to the foot of the object, when both stand on the same level ground.
• Angle of Elevation — the angle formed by the line of sight with the horizontal line, when the object being viewed is above the horizontal level (the observer has to look upward).
• Angle of Depression — the angle formed by the line of sight with the horizontal line, when the object being viewed is below the horizontal level (the observer has to look downward).
• As the observer or the object moves closer to the other, the angle of elevation/depression increases; as they move farther apart, it decreases.

Tips for Solving Problems

• Read the question carefully and draw a clear, labelled figure before starting to solve.
• Identify the side that is common to two right triangles formed in a question; if its value is unknown, find it first.
• Use whichever trigonometric ratio is convenient — for example, using tan A instead of cot A gives the same correct answer.
• If a side is split into two parts and its total length is known, take one part as x and the other as (total length − x).
• The angle of depression from one point to an object equals the angle of elevation from the object to that point, since the two horizontal lines are parallel (alternate angles).

Worked Examples

Example 1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Solution: Let the height of the pole = AB = h metres, and the length of the rope = AC = 20 m, with ∠C = 30°.

In right triangle ABC: sin 30° = AB/AC

1/2 = h/20 ⟹ h = 10

Hence, the height of the pole is 10 m.

Example 2

A tree breaks due to a storm, and the broken part bends so that the top of the tree touches the ground, making an angle of 30° with the ground. The distance between the foot of the tree and the point where the top touches the ground is 8 m. Find the height of the tree.

Solution: Let the tree originally be AB, breaking at point C. BC remains standing vertically, and the broken part CA bends to touch the ground at A′, with ∠A′ = 30° and BA′ = 8 m.

In right triangle BCA′: tan 30° = BC/BA′ ⟹ BC = 8 × (1/√3) = 8/√3 m

Also, sin 30° = BC/CA′ ⟹ CA′ = BC/sin 30° = (8/√3)/(1/2) = 16/√3 m

Total height of the tree = BC + CA′ = 8/√3 + 16/√3 = 24/√3 = 8√3 m

Hence, the height of the tree is 8√3 m (≈ 13.86 m).

Example 3

A contractor wants to set up a slide for children in a park. For children below 5 years, she wants to build a slide whose top is at a height of 1.5 m and is inclined at 30° to the ground. For older children, she wants a steeper slide at a height of 3 m, inclined at 60° to the ground. Find the length of the slide in each case.

Solution: Let the length of the slide = L (the hypotenuse), and the height = the perpendicular side.

Case 1 (height 1.5 m, 30°): sin 30° = 1.5/L ⟹ L = 1.5/(1/2) = 3 m

Case 2 (height 3 m, 60°): sin 60° = 3/L ⟹ L = 3/(√3/2) = 6/√3 = 2√3 m (≈ 3.46 m)

Example 4

From a point on the ground, 30 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

Solution: Let the height of the tower = h m. tan 30° = h/30

1/√3 = h/30 ⟹ h = 30/√3 = 10√3 m (≈ 17.32 m)

Example 5

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in it.

Solution: sin 60° = 60/L ⟹ L = 60/(√3/2) = 120/√3 = 40√3 m (≈ 69.28 m)

Example 6

A boy who is 1.5 m tall is standing at some distance from a 30 m tall building. As he walks towards the building, the angle of elevation of the top of the building from his eye changes from 30° to 60°. Find the distance he walked towards the building.

Solution: Height of the building above the boy’s eye level = 30 − 1.5 = 28.5 m.

Initial distance: tan 30° = 28.5/d₁ ⟹ d₁ = 28.5√3

Final distance: tan 60° = 28.5/d₂ ⟹ d₂ = 28.5/√3 = 9.5√3

Distance walked = d₁ − d₂ = 28.5√3 − 9.5√3 = 19√3 m (≈ 32.91 m)

Example 7

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution: Let the height of the tower = h m, and the distance of the point from the foot of the building = d.

For the building top (bottom of the tower): tan 45° = 20/d ⟹ d = 20 m

For the top of the tower: tan 60° = (20 + h)/d ⟹ √3 = (20 + h)/20 ⟹ 20 + h = 20√3

h = 20√3 − 20 = 20(√3 − 1) m (≈ 14.64 m)

Example 8

A statue, 1.6 m tall, stands on top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60°, and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution: Let the height of the pedestal = h m, and its distance from the point = d.

tan 45° = h/d ⟹ d = h

tan 60° = (h + 1.6)/d = (h + 1.6)/h = √3 ⟹ h + 1.6 = h√3

h(√3 − 1) = 1.6 ⟹ h = 1.6/(√3 − 1) = 1.6(√3 + 1)/2 = 0.8(√3 + 1) m (≈ 2.19 m)

Example 9

The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution: Let the height of the building = h m, and the distance between them = d.

tan 60° = 50/d ⟹ d = 50/√3

tan 30° = h/d ⟹ h = d × (1/√3) = (50/√3) × (1/√3) = 50/3 m = 16.67 m

Example 10

Two poles of equal height stand on either side of a road, 80 m wide. From a point between them on the road, the angles of elevation of the tops of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Solution: Let the height of the poles = h m; distance of the point from one pole = x m, and from the other = (80 − x) m.

tan 60° = h/x ⟹ h = x√3

tan 30° = h/(80 − x) ⟹ h = (80 − x)/√3

Equating: x√3 = (80 − x)/√3 ⟹ 3x = 80 − x ⟹ 4x = 80 ⟹ x = 20

h = 20√3 m (≈ 34.64 m). Hence, the height of each pole is 20√3 m, and the point is 20 m and 60 m from the two poles respectively.

Example 11

A TV tower stands vertically on the bank of a canal. From a point on the other bank, directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point, on the same bank and in line with the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Solution: Let the height of the tower = h m, and the width of the canal = x m.

tan 60° = h/x ⟹ h = x√3

tan 30° = h/(x + 20) ⟹ h = (x + 20)/√3

Equating: x√3 = (x + 20)/√3 ⟹ 3x = x + 20 ⟹ 2x = 20 ⟹ x = 10

h = 10√3 m. Hence, the width of the canal is 10 m and the height of the tower is 10√3 m (≈ 17.32 m).

Example 12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60°, and the angle of depression of its foot is 45°. Find the height of the tower.

Solution: Let the height of the tower = h m, and the horizontal distance between the building and the tower = d.

Angle of depression to the foot of the tower = 45° ⟹ tan 45° = 7/d ⟹ d = 7 m

Angle of elevation to the top of the tower (rise above the building’s top level) = 60°:

tan 60° = (h − 7)/d = (h − 7)/7 = √3 ⟹ h − 7 = 7√3 ⟹ h = 7 + 7√3 = 7(1 + √3) m (≈ 19.12 m)

Example 13

As observed from the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is directly behind the other, on the same side of the lighthouse, find the distance between the two ships.

Solution: Let the distances of the ships from the base of the lighthouse be d₁ (for 45°) and d₂ (for 30°).

tan 45° = 75/d₁ ⟹ d₁ = 75 m

tan 30° = 75/d₂ ⟹ d₂ = 75√3 m

Distance between the ships = d₂ − d₁ = 75√3 − 75 = 75(√3 − 1) m (≈ 54.9 m)

Example 14

A girl 1.2 m tall spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. At a given instant, the angle of elevation of the balloon from her eyes is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during this interval.

Solution: Height of the balloon above the girl’s eye level = 88.2 − 1.2 = 87 m.

At 60°: tan 60° = 87/d₁ ⟹ d₁ = 87/√3 = 29√3 m

At 30°: tan 30° = 87/d₂ ⟹ d₂ = 87√3 m

Distance travelled = d₂ − d₁ = 87√3 − 29√3 = 58√3 m (≈ 100.46 m)

Example 15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, moving towards the tower with uniform speed. Six seconds later, the angle of depression becomes 60°. Find the time taken by the car to reach the foot of the tower from this second point.

Solution: Let the height of the tower = h m. Distance of the car at 30° = h√3; at 60° = h/√3.

Distance covered in 6 seconds = h√3 − h/√3 = (3h − h)/√3 = 2h/√3

Speed = (2h/√3) ÷ 6 = h/(3√3)

Time to cover the remaining distance (h/√3) at this speed = (h/√3) ÷ (h/(3√3)) = (h/√3) × (3√3/h) = 3

Hence, the car takes a further 3 seconds to reach the foot of the tower.

Example 16

The angles of elevation of the top of a tower, from two points at distances of 4 m and 9 m from the base of the tower, on the same straight line, are complementary. Prove that the height of the tower is 6 m.

Solution: Let the height of the tower = h m, and let the two angles be α and β, with α + β = 90°.

tan α = h/4 and tan β = h/9

Since β = 90° − α, tan β = cot α = 1/tan α, so 1/tan α = h/9 ⟹ tan α = 9/h

Also, tan α = h/4. Equating: h/4 = 9/h ⟹ h² = 36 ⟹ h = 6

Hence, the height of the tower is proved to be 6 m.

Key Points to Remember

• Angle of elevation is measured upward from the horizontal line; angle of depression is measured downward from the horizontal line.
• The angle of depression from an observer’s position equals the angle of elevation from the object being observed (alternate angles).
• Always draw and label a clear right-triangle figure before solving a heights-and-distances problem.
• tan θ = height/base distance is the ratio most commonly used to relate height and horizontal distance.
• When two angles are given at different points along the same line, form two equations using tan θ and solve them together.
• Take a person’s eye-level height into account whenever it is given in the problem.
• Keep answers in surd (√) form unless the question specifically asks for a decimal approximation.

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