Applications of Trigonometry - Solutions

Applications of Trigonometry - 1 Mark Solutions

Important Trigonometric Concepts:

• Angle of Elevation: Angle between horizontal line and line of sight when looking up
• Angle of Depression: Angle between horizontal line and line of sight when looking down
• tan θ = Opposite/Adjacent
• Height = Distance × tan(angle of elevation)
• When object and shadow are equal, angle = 45°
• As angle increases from 0° to 90°, tan θ increases from 0 to ∞

A person from the top of a building of height 25m has observed another building's top and bottom at an angle of elevation of 45° and at an angle of depression 60° respectively. Draw a diagram for this data. (M'15)

Step 1: Understand the scenario

- First building height: 25m
- Observer at top of first building
- Angle of elevation to top of second building: 45°
- Angle of depression to bottom of second building: 60°

Step 2: Diagram description

The diagram shows:
- Two vertical buildings with some distance between them
- First building of height 25m
- Observer at top of first building
- Line of sight to top of second building at 45° (angle of elevation)
- Line of sight to bottom of second building at 60° (angle of depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer at 25m height with lines of sight at 45° elevation and 60° depression.

A person observed the top of a tree at an angle of elevation of 60° when the observation point was 5m away from the foot of the tree. Draw a diagram for this data. (J'15)

Step 1: Understand the scenario

- Distance from tree: 5m
- Angle of elevation: 60°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tree
- Observer at ground level, 5m from tree
- Line of sight to top of tree at 60° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 5m from tree with 60° angle of elevation to the top.

"If the angle of elevation of Sun increases from 0° to 90°, then the length of a tower decreases", is this statement true? Justify your answer. (M'16)

Step 1: Understand the relationship

Let:
- h = height of tower (constant)
- θ = angle of elevation of sun
- l = length of shadow

Step 2: Apply trigonometry

tan θ = h/l
So, l = h / tan θ = h × cot θ

Step 3: Analyze the behavior

As θ increases from 0° to 90°:
- tan θ increases from 0 to ∞
- cot θ decreases from ∞ to 0
Therefore, l = h × cot θ decreases from ∞ to 0

Shadow Length Calculator

Tower Height (m):

Sun Angle (degrees):

∴ Yes, the statement is true. As the sun's angle increases from 0° to 90°, the shadow length decreases from infinity to zero.

If a tower of height 'h' is observed from a point with a distance 'd' and angle 'θ', then express the relation among h, d and θ. (J'16)

Step 1: Visualize the scenario

- Tower height = h
- Distance from tower = d
- Angle of elevation = θ

Step 2: Apply trigonometric ratio

In the right triangle formed:
- Opposite side = h (tower height)
- Adjacent side = d (distance)
- Angle = θ

Step 3: Write the relation

tan θ = Opposite/Adjacent = h/d
Therefore, h = d × tan θ

Diagram:

∴ The relation is h = d × tan θ

A pole and its shadow have same length, find the angle of the ray made with the earth at that time. (M'18)

Step 1: Understand the scenario

- Pole height = h
- Shadow length = h (same as pole height)
- We need to find the angle θ that the sun's rays make with the ground

Step 2: Apply trigonometry

In the right triangle formed:
- Opposite side = h (pole height)
- Adjacent side = h (shadow length)
tan θ = Opposite/Adjacent = h/h = 1

Step 3: Find the angle

tan θ = 1
We know that tan 45° = 1
Therefore, θ = 45°

Diagram:

∴ The angle of the sun's rays with the earth is 45°.

An observer observes the top of a tower from two points on the same side of a tower and on the same straight line which are at distances of 5 m and 8 m with angles of elevation 60° and 45° respectively. Draw a suitable diagram for the given data. (Aug.22)

Step 1: Understand the scenario

- Tower of unknown height
- First observation point: 5m from tower, angle = 60°
- Second observation point: 8m from tower, angle = 45°
- Both points on same side of tower and in straight line

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Two observation points at distances 5m and 8m from tower
- Lines of sight from both points to top of tower
- Angles of elevation marked as 60° and 45°

Diagram:

∴ The diagram shows two observation points at 5m and 8m with elevation angles 60° and 45° respectively.

"An observer standing at a distance of 10m from the foot of a tower, observes its top with an angle of elevation of 60°". Draw a suitable diagram for this situation. (Apr'.23)

Step 1: Understand the scenario

- Distance from tower: 10m
- Angle of elevation: 60°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Observer at ground level, 10m from tower
- Line of sight to top of tower at 60° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 10m from tower with 60° angle of elevation to the top.

Applications of Trigonometry (2 Marks) - Complete Solutions

Important Trigonometric Concepts:

A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to stream of the river and travels a distance of 450m to reach the another side of the river. Draw the diagram for this data. (M'16)

Step 1: Understand the scenario

- Boat makes 60° angle with river bank
- Distance traveled: 450m
- Due to stream, boat doesn't travel perpendicular to bank

Step 2: Diagram description

The diagram shows:
- Two parallel lines representing river banks
- Boat starting from one bank at 60° angle
- Actual path of boat (450m at 60°)
- Perpendicular distance between banks (width of river)
- Components: actual path and perpendicular crossing

Diagram:

∴ The diagram shows the boat crossing at 60° angle due to river stream.

A person 25 m away from a cell tower observes the top of the cell tower at an angle of elevation 30°. Draw the suitable diagram for this situation. (M'17)

Step 1: Understand the scenario

- Distance from tower: 25m
- Angle of elevation: 30°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical cell tower
- Observer at ground level, 25m from tower
- Line of sight to top of tower at 30° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 25m from cell tower with 30° angle of elevation.

A State highway leads to foot of the tower. A Man Standing at the top of the tower observes a car at an angle of depression of θ, which is approaching to the foot of the tower with a uniform speed. 6 seconds later the angle of depression is φ. Draw a diagram for this data and analyze. (J'17)

Step 1: Understand the scenario

- Tower of height h
- Car moving towards tower on highway
- Initial angle of depression: θ
- After 6 seconds, angle of depression: φ
- φ > θ (as car gets closer, angle increases)
- Car moving with uniform speed

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Highway leading to foot of tower
- Observer at top of tower
- Two positions of car: initial and after 6 seconds
- Lines of sight with angles θ and φ
- Distances from tower: d₁ and d₂ (d₁ > d₂)

Diagram:

Step 3: Analysis

Let:
- Tower height = h
- Initial distance = d₁
- Final distance = d₂
- Speed of car = v

From trigonometry:
tan θ = h/d₁ ⇒ d₁ = h/tan θ
tan φ = h/d₂ ⇒ d₂ = h/tan φ

Distance covered in 6 seconds = d₁ - d₂ = h(1/tan θ - 1/tan φ)
Speed v = (d₁ - d₂)/6 = h(1/tan θ - 1/tan φ)/6

∴ The diagram shows car approaching tower with angles θ and φ. Speed can be calculated using tower height and angles.

From the top of the tower of height h m height, Anusha observes the angles of depression of two points X and Y on the same side of the tower on the ground to be α and β. Draw the suitable figure for the given information. (M'18)

Step 1: Understand the scenario

- Tower height = h
- Two points X and Y on same side of tower
- Angles of depression: α and β
- Assume α > β (point X is closer than Y)

Step 2: Diagram description

The diagram shows:
- A vertical tower of height h
- Observer at top of tower
- Two points X and Y on ground
- Lines of sight to X and Y with angles α and β
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows Anusha observing points X and Y with angles of depression α and β.

The angle of elevation of the top of a tower from a point on the ground, which is 50m away from the foot of the tower, is 45°. Draw the diagram for the situation. (J'18)

Step 1: Understand the scenario

- Distance from tower: 50m
- Angle of elevation: 45°
- Observer at ground level

Step 2: Diagram description

The diagram shows:
- A vertical tower
- Observer at ground level, 50m from tower
- Line of sight to top of tower at 45° angle
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows the observer 50m from tower with 45° angle of elevation.

From the top of the building the angle of elevation of the top of the cell tower is 60° and the angle of depression to its foot is 45°, if the distance of the building from the tower is 30m, draw the suitable diagram to the given data. (M'19)

Step 1: Understand the scenario

- Distance between building and tower: 30m
- Observer at top of building
- Angle of elevation to top of tower: 60°
- Angle of depression to foot of tower: 45°

Step 2: Diagram description

The diagram shows:
- Two vertical structures: building and cell tower
- Distance between them: 30m
- Observer at top of building
- Line of sight to top of tower at 60° (elevation)
- Line of sight to foot of tower at 45° (depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows observer on building with 60° elevation to tower top and 45° depression to tower foot.

From the top of the building, The angle of elevation of the top of a TV tower is α° and the angle of depression to its (T.V. tower) foot is β°. If distance of the building from the tower is 'd' metres, draw the suitable diagram of the given data. (J'19)

Step 1: Understand the scenario

- Distance between building and tower: d meters
- Observer at top of building
- Angle of elevation to top of TV tower: α°
- Angle of depression to foot of TV tower: β°

Step 2: Diagram description

The diagram shows:
- Two vertical structures: building and TV tower
- Distance between them: d meters
- Observer at top of building
- Line of sight to top of TV tower at α° (elevation)
- Line of sight to foot of TV tower at β° (depression)
- Horizontal line from observer's eye level

Diagram:

∴ The diagram shows observer on building with α° elevation to TV tower top and β° depression to tower foot.

The statue stands on the top of 3 m tall pedestal. From a point on the ground 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 statue. (Aug.22)

Step 1: Understand the scenario

- Pedestal height = 3m
- Statue height = ? (let's call it h)
- Same observation point for both measurements
- Angle to top of statue = 60°
- Angle to top of pedestal = 45°

Step 2: Diagram description

The diagram shows:
- Pedestal of height 3m
- Statue of height h on top of pedestal
- Observer at ground level
- Lines of sight to top of pedestal (45°) and top of statue (60°)
- Distance from observer to pedestal: d

Diagram:

Step 3: Set up equations

Let:
- Distance from observer = d
- Pedestal height = 3m
- Statue height = h

For pedestal top (45°):
tan 45° = 3/d ⇒ 1 = 3/d ⇒ d = 3m

For statue top (60°):
tan 60° = (3 + h)/d ⇒ √3 = (3 + h)/3

Step 4: Solve for statue height

√3 = (3 + h)/3
3√3 = 3 + h
h = 3√3 - 3
h = 3(√3 - 1)

Using √3 ≈ 1.732:
h = 3(1.732 - 1) = 3 × 0.732 = 2.196m ≈ 2.2m

Calculate Statue Height

Pedestal Height (m):

Angle to Pedestal (°):

Angle to Statue (°):

∴ The height of the statue is 3(√3 - 1) m ≈ 2.2 m.

Applications of Trigonometry - 4 Marks Solutions

Important Trigonometric Concepts:

An observer flying in an altitude of 900m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships. (M'15)

Step 1: Understand the scenario

- Observer's altitude: 900m
- Angle of depression to first ship: 60°
- Angle of depression to second ship: 30°
- Both ships are in the same direction

Step 2: Apply trigonometry

Let the distance from observer's vertical to first ship = d₁
Let the distance from observer's vertical to second ship = d₂
tan(60°) = 900/d₁ ⇒ d₁ = 900/tan(60°) = 900/√3 = 300√3 m
tan(30°) = 900/d₂ ⇒ d₂ = 900/tan(30°) = 900/(1/√3) = 900√3 m

Step 3: Find distance between ships

Distance between ships = d₂ - d₁ = 900√3 - 300√3 = 600√3 m

∴ The distance between the two ships is 600√3 m ≈ 1039.23 m

A person from the top of a building of height 15 meters observes the top and the bottom of a cell tower with the angle of elevation as 60° and the angle of depression as 45° respectively. Then find the height of the cell tower. (J'15)

Step 1: Understand the scenario

- Building height: 15m
- Angle of elevation to top of tower: 60°
- Angle of depression to bottom of tower: 45°
- We need to find height of cell tower

Step 2: Apply trigonometry

Let distance between buildings = d
From angle of depression 45°: tan(45°) = 15/d ⇒ d = 15m
Let height of tower above observer = h
From angle of elevation 60°: tan(60°) = h/d ⇒ h = d × tan(60°) = 15 × √3 = 15√3 m

Step 3: Find total height of tower

Total height of tower = height below observer + height above observer
= 15 + 15√3 = 15(1 + √3) m

∴ The height of the cell tower is 15(1 + √3) m ≈ 40.98 m

Two poles of equal heights are standing opposite to each other, on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30° respectively. Find the height of the poles. (M'16)

Step 1: Understand the scenario

- Road width: 80m
- Two poles of equal height h
- Angles of elevation: 60° and 30°
- Observation point is between them on the road

Step 2: Apply trigonometry

Let distance from observation point to first pole = x
Then distance to second pole = 80 - x
For first pole: tan(60°) = h/x ⇒ h = x√3
For second pole: tan(30°) = h/(80 - x) ⇒ h = (80 - x)/√3

Step 3: Solve for x and h

Equating both expressions for h:
x√3 = (80 - x)/√3
3x = 80 - x
4x = 80 ⇒ x = 20m
Then h = x√3 = 20√3 m

∴ The height of each pole is 20√3 m ≈ 34.64 m

A tree is broken without separating from the stem by the wind. The top touches the ground making an angle 30° at a distance of 12m from the foot of the tree. Find the height of the tree before breaking. (J'16)

Step 1: Understand the scenario

- Tree broken but still attached to stem
- Top touches ground at 30° angle
- Distance from foot to where top touches: 12m
- Need to find original height of tree

Step 2: Apply trigonometry

Let height of standing part = h₁
Let length of broken part = h₂
In the triangle formed by broken part:
cos(30°) = 12/h₂ ⇒ h₂ = 12/cos(30°) = 12/(√3/2) = 24/√3 = 8√3 m
sin(30°) = h₁/12 ⇒ h₁ = 12 × sin(30°) = 12 × 1/2 = 6m

Step 3: Find total height

Total height = h₁ + h₂ = 6 + 8√3 m

∴ The original height of the tree was 6 + 8√3 m ≈ 19.86 m

Two poles are standing opposite to each other on the either side of the road which is 90 feet wide. The angle of elevation from bottom of the first pole to the top of the second pole is 45°. The angle of elevation from the bottom of the second pole to the top of the first pole is 30°. Find the heights of the poles. (use √3 = 1.732) (M'17)

Step 1: Understand the scenario

- Road width: 90 feet
- Two poles of different heights h₁ and h₂
- Angle from bottom of first to top of second: 45°
- Angle from bottom of second to top of first: 30°

Step 2: Apply trigonometry

For first observation: tan(45°) = h₂/90 ⇒ h₂ = 90 × 1 = 90 feet
For second observation: tan(30°) = h₁/90 ⇒ h₁ = 90 × 1/√3 = 90/√3 = 30√3 feet

Step 3: Calculate numerical values

h₁ = 30√3 ≈ 30 × 1.732 = 51.96 feet
h₂ = 90 feet

∴ The heights of the poles are approximately 51.96 feet and 90 feet

The angle of elevation of top of the tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it, are complementary. Prove that the height of the tower is 6 m. (J'17)

Step 1: Understand the scenario

- Tower height: h
- Two observation points at 4m and 9m from base
- Angles of elevation: θ and (90° - θ) [complementary]
- Need to prove h = 6m

Step 2: Apply trigonometry

From first point: tan θ = h/4
From second point: tan(90° - θ) = h/9 ⇒ cot θ = h/9
Since tan θ × cot θ = 1
(h/4) × (h/9) = 1
h²/36 = 1 ⇒ h² = 36 ⇒ h = 6m

∴ The height of the tower is 6 m, as required to prove.

From the top of a tower of 50m high, Neha observes the angles of depression of the top and foot of another building to be 45° and 60° respectively. Find the height of the building. (M'18)

Step 1: Understand the scenario

- First tower height: 50m
- Angle of depression to top of building: 45°
- Angle of depression to foot of building: 60°
- Need to find height of building

Step 2: Apply trigonometry

Let distance between buildings = d
From angle of depression to foot: tan(60°) = 50/d ⇒ d = 50/√3 m
Let height difference between towers = h
From angle of depression to top: tan(45°) = h/d ⇒ h = d = 50/√3 m

Step 3: Find building height

Building height = 50 - h = 50 - 50/√3 = 50(1 - 1/√3) m

∴ The height of the building is 50(1 - 1/√3) m ≈ 21.13 m

Two boys on either side of their school building of 20 m height observes its top at the angles of elevation 30° and 60° respectively. Find the distance between two boys. (J'18)

Step 1: Understand the scenario

- Building height: 20m
- Angles of elevation: 30° and 60°
- Boys on opposite sides of building
- Need to find distance between boys

Step 2: Apply trigonometry

For first boy: tan(30°) = 20/d₁ ⇒ d₁ = 20/tan(30°) = 20√3 m
For second boy: tan(60°) = 20/d₂ ⇒ d₂ = 20/tan(60°) = 20/√3 m

Step 3: Find total distance

Distance between boys = d₁ + d₂ = 20√3 + 20/√3
= 20(√3 + 1/√3) = 20(3 + 1)/√3 = 80/√3 m

∴ The distance between the two boys is 80/√3 m ≈ 46.19 m

The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high. Find the height of the hill. (M'19)

Step 1: Understand the scenario

- Tower height: 50m
- Angle from tower base to hill top: 60°
- Angle from hill base to tower top: 30°
- Need to find hill height

Step 2: Apply trigonometry

Let distance between tower and hill = d
From tower: tan(30°) = 50/d ⇒ d = 50/tan(30°) = 50√3 m
Let hill height = h
From hill: tan(60°) = h/d ⇒ h = d × tan(60°) = 50√3 × √3 = 50 × 3 = 150m

∴ The height of the hill is 150 m

A man observes top of tower at an angle of elevation of 30°. When he walked 40 m towards the tower, the angle of elevation is changed to 60°. Find the height of the tower and distance from the first observation point to the tower. (J'19)

Step 1: Understand the scenario

- First angle of elevation: 30°
- After walking 40m towards tower: angle becomes 60°
- Need to find tower height and initial distance

Step 2: Apply trigonometry

Let tower height = h
Let initial distance = d
From first point: tan(30°) = h/d ⇒ h = d/√3
From second point: tan(60°) = h/(d - 40) ⇒ h = (d - 40)√3
Equating: d/√3 = (d - 40)√3
d = 3(d - 40) ⇒ d = 3d - 120 ⇒ 2d = 120 ⇒ d = 60m

Step 3: Find tower height

h = d/√3 = 60/√3 = 20√3 m

∴ The height of the tower is 20√3 m ≈ 34.64 m and the initial distance was 60 m

If two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60° and 45° respectively, then find the distance between the two persons. (May 2022)

Step 1: Understand the scenario

- Tower height: 100m
- Angles of elevation: 60° and 45°
- Persons on opposite sides of tower
- Need to find distance between persons

Step 2: Apply trigonometry

For first person: tan(60°) = 100/d₁ ⇒ d₁ = 100/tan(60°) = 100/√3 m
For second person: tan(45°) = 100/d₂ ⇒ d₂ = 100/tan(45°) = 100/1 = 100m

Step 3: Find total distance

Distance between persons = d₁ + d₂ = 100/√3 + 100 = 100(1 + 1/√3) m

∴ The distance between the two persons is 100(1 + 1/√3) m ≈ 157.74 m

If two boys standing on either side of their school building of height 20m, observed the top of it with angles of elevation of 30° and 60° respectively, then find the distance between the two boys. (Jun'23)

Step 1: Understand the scenario

- Building height: 20m
- Angles of elevation: 30° and 60°
- Boys on opposite sides of building
- Need to find distance between boys

Step 2: Apply trigonometry

For first boy: tan(30°) = 20/d₁ ⇒ d₁ = 20/tan(30°) = 20√3 m
For second boy: tan(60°) = 20/d₂ ⇒ d₂ = 20/tan(60°) = 20/√3 m

Step 3: Find total distance

Distance between boys = d₁ + d₂ = 20√3 + 20/√3
= 20(√3 + 1/√3) = 20(3 + 1)/√3 = 80/√3 m

∴ The distance between the two boys is 80/√3 m ≈ 46.19 m

Author

zaheerpashamd

M.Sc,B.Ed: SDVR ZPHS B.Gangaram Sathupally(md) Khammam(dt)
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