Some Applications of Trigonometry | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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10th

10th - Mathematics

Some Applications of Trigonometry

Overview of the Chapter

This chapter, "Some Applications of Trigonometry," introduces students to the practical uses of trigonometric concepts in real-world scenarios. It focuses on solving problems related to heights and distances using trigonometric ratios, angle of elevation, and angle of depression.

Angle of Elevation: The angle formed by the line of sight with the horizontal when the object being viewed is above the horizontal level.

Angle of Depression: The angle formed by the line of sight with the horizontal when the object being viewed is below the horizontal level.

Key Concepts

Understanding the line of sight, angle of elevation, and angle of depression.
Using trigonometric ratios (sin, cos, tan) to find heights and distances.
Solving real-life problems involving heights of towers, buildings, or other objects.

Applications

Trigonometry is widely used in fields such as astronomy, navigation, engineering, and architecture. This chapter helps students apply trigonometric principles to measure inaccessible distances and heights.

Example Problems

Finding the height of a tower when the angle of elevation and distance from the tower are given.
Calculating the distance between two objects using angles of elevation or depression.

Summary

This chapter bridges theoretical trigonometry with practical applications, enhancing problem-solving skills and logical thinking. Students learn to model real-world situations mathematically using trigonometric concepts.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

A ladder leans against a wall making a 60° angle with the ground. If the foot is 2m away, find the ladder's length.

Answer:

4 m

Question 2:

The angle of elevation of the sun is 45°. A tree casts a 6m shadow. Find its height.

Answer:

6 m

Question 3:

A kite flies at a height of 60m with 75m string. Find the angle of elevation.

Answer:

53°

Question 4:

From a tower 20m high, the angle of depression of a car is 30°. Find its distance from the tower.

Answer:

20√3 m

Question 5:

A pole’s shadow is √3 times its height. Find the sun’s angle of elevation.

Answer:

30°

Question 6:

A 12m ladder makes a 30° angle with a wall. How far is its foot from the wall?

Answer:

6 m

Question 7:

The angle of elevation changes from 30° to 60° as an observer moves 20m closer. Find initial distance.

Answer:

30 m

Question 8:

A 1.5m tall boy sees a bird at 45° elevation. If he’s 15m away, find the bird’s height.

Answer:

16.5 m

Question 9:

A tower’s shadow is 10m long when the sun’s angle of elevation is 30°. Find the tower’s height.

Answer:

10/√3 m

Question 10:

From a boat, a 100m cliff’s angle of elevation is 45°. Find the boat’s distance from the cliff.

Answer:

100 m

Question 11:

Two poles of heights 6m and 11m are 10m apart. Find the angle of elevation of the top of one from the other.

Answer:

45°

Question 12:

A 5m long ladder reaches a window 4m high. Find the ladder’s angle of inclination with the ground.

Answer:

53°

Question 13:

A ladder leans against a wall making a 60° angle with the ground. If the ladder is 10m long, how high does it reach?

Answer:

8.66 m
Using sin 60° = height/10.

Question 14:

The angle of elevation of the top of a tower is 30° from 30m away. Find its height.

Answer:

17.32 m
tan 30° = height/30.

Question 15:

A kite flies at a height of 60m with 80m string. What angle does the string make with the ground?

Answer:

48.59°
sin⁻¹(60/80).

Question 16:

From a point 20m away, the angle of elevation to a tree's top is 45°. Find the tree's height.

Answer:

20 m
tan 45° = height/20.

Question 17:

A 12m shadow is cast when the sun's elevation is 30°. Find the object's height.

Answer:

6.93 m
tan 30° = height/12.

Question 18:

A pole's shadow is √3 times its height. Find the sun's angle of elevation.

Answer:

30°
tan θ = height/(√3 × height).

Question 19:

The angle of depression from a 15m tower to a point is 30°. How far is the point?

Answer:

25.98 m
tan 30° = 15/distance.

Question 20:

A 1.5m tall observer sees a cloud at 60° elevation. If the cloud is 100m high, how far is it?

Answer:

56.89 m
tan 60° = (100-1.5)/distance.

Question 21:

Two poles of heights 6m and 11m stand 12m apart. Find the distance between their tops.

Answer:

13 m
√(5² + 12²) using Pythagoras.

Question 22:

A ship sees a lighthouse at 45° elevation 100m away. What is the lighthouse height?

Answer:

100 m
tan 45° = height/100.

Question 23:

The ratio of a building's height to its shadow is 1:√3. Find the sun's elevation.

Answer:

30°
tan θ = 1/√3.

Question 24:

From a cliff 50m high, the angle of depression to a boat is 45°. How far is the boat?

Answer:

50 m
tan 45° = 50/distance.

Question 25:

A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 m away from the wall, find the length of the ladder.

Answer:

Using cosine of the angle:
cos(60°) = Adjacent side / Hypotenuse
1/2 = 2 / Length of ladder
Length of ladder = 4 m

Question 26:

The angle of elevation of the top of a tower from a point on the ground is 30°. If the point is 30 m away from the base of the tower, find the height of the tower.

Answer:

Using tangent of the angle:
tan(30°) = Height / Distance
1/√3 = Height / 30
Height = 30/√3 = 10√3 m

Question 27:

A kite is flying at a height of 60 m with a string inclined at 60° to the horizontal. Find the length of the string.

Answer:

Using sine of the angle:
sin(60°) = Height / Length of string
√3/2 = 60 / Length of string
Length of string = 120/√3 = 40√3 m

Question 28:

The shadow of a vertical pole is √3 times its height. Find the angle of elevation of the sun.

Answer:

Let height = h, shadow = h√3
tanθ = Height / Shadow
tanθ = h / (h√3) = 1/√3
θ = 30°

Question 29:

From the top of a 10 m high building, the angle of depression of a point on the ground is 30°. Find the distance of the point from the base of the building.

Answer:

Using tangent of the angle:
tan(30°) = Height / Distance
1/√3 = 10 / Distance
Distance = 10√3 m

Question 30:

A tree breaks due to a storm and the broken part bends so that the top touches the ground at an angle of 30°. If the distance from the foot to the point where the top touches the ground is 5 m, find the original height of the tree.

Answer:

Using cosine and Pythagoras theorem:
cos(30°) = 5 / Length of broken part
√3/2 = 5 / Length ⇒ Length = 10/√3 m
Height of standing part = √[(10/√3)² - 5²] = 5/√3 m
Total height = (10/√3 + 5/√3) = 15/√3 = 5√3 m

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 m away from the wall, find the length of the ladder.

Answer:

Let the length of the ladder be L.
Given: Base = 2 m, Angle = 60°.
Using cosine of the angle:
cos 60° = Base / Hypotenuse
1/2 = 2 / L
L = 4 m.
Thus, the ladder is 4 meters long.

Question 2:

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

Answer:

Let the height of the tower be h.
Given: Distance = 30 m, Angle = 30°.
Using tangent of the angle:
tan 30° = Height / Distance
1/√3 = h / 30
h = 30/√3 = 10√3 m.
Thus, the tower is 10√3 meters tall.

Question 3:

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. If the inclination of the string with the ground is 60°, find the length of the string.

Answer:

Let the length of the string be L.
Given: Height = 60 m, Angle = 60°.
Using sine of the angle:
sin 60° = Height / Hypotenuse
√3/2 = 60 / L
L = 120/√3 = 40√3 m.
Thus, the string is 40√3 meters long.

Question 4:

From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.

Answer:

Let the height of the tower be H.
Distance between building and tower = 10 m (since angle of depression is 45°).
Using tangent for elevation:
tan 60° = (H - 10) / 10
√3 = (H - 10) / 10
H = 10√3 + 10 = 10(√3 + 1) m.
Thus, the tower is 10(√3 + 1) meters tall.

Question 5:

A flagpole casts a shadow of 15 m when the angle of elevation of the sun is 30°. Find the height of the flagpole.

Answer:

Let the height of the flagpole be h.
Given: Shadow length = 15 m, Angle = 30°.
Using tangent of the angle:
tan 30° = Height / Shadow
1/√3 = h / 15
h = 15/√3 = 5√3 m.
Thus, the flagpole is 5√3 meters tall.

Question 6:

The angle of elevation of the top of a hill from a point on the ground is 45°. After walking 100 m towards the hill, the angle of elevation becomes 60°. Find the height of the hill.

Answer:

Let the height of the hill be h.
Initial distance = d, New distance = d - 100.
Using tangent for both angles:
tan 45° = h / d ⇒ h = d.
tan 60° = h / (d - 100) ⇒ √3 = h / (h - 100).
Solving: h = 100√3 / (√3 - 1) ≈ 236.6 meters (rationalized).

Question 7:

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

Answer:

Let the original height be H = Standing part (h) + Broken part (L).
Given: Base = 8 m, Angle = 30°.
Using trigonometry:
tan 30° = h / 8 ⇒ h = 8/√3.
cos 30° = 8 / L ⇒ L = 16/√3.
H = h + L = 24/√3 = 8√3 m.
Thus, the tree was 8√3 meters tall.

Question 8:

Two poles of heights 10 m and 15 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Answer:

The difference in heights = 15 - 10 = 5 m.
Horizontal distance = 12 m.
Using the Pythagoras theorem:
Distance between tops = √(5² + 12²)
= √(25 + 144)
= √169 = 13 meters.

Question 9:

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Answer:

Using the Pythagoras theorem in the right triangle formed by the ladder, wall, and ground:
Let the distance from the foot of the ladder to the wall be x.
Then, x² + 8² = 10²
x² + 64 = 100
x² = 36
x = 6 m
Thus, the distance is 6 meters.

Question 10:

The shadow of a vertical pole is √3 times its height. Find the angle of elevation of the sun at that time.

Answer:

Let the height of the pole be h and the angle of elevation be θ.
The shadow length is √3 h.
Using the tangent ratio: tan θ = h / (√3 h) = 1/√3
Thus, θ = 30° (since tan 30° = 1/√3).
The angle of elevation is 30°.

Question 11:

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

Answer:

Let the height of the tower be H and the distance between the building and the tower be x.
Using the tangent ratio for the angle of depression (45°):
tan 45° = 7 / x ⇒ x = 7 m
Now, for the angle of elevation (60°):
tan 60° = (H - 7) / x ⇒ √3 = (H - 7) / 7
H - 7 = 7√3 ⇒ H = 7 + 7√3 m
Thus, the height of the tower is 7(1 + √3) meters.

Question 12:

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 it. If the distance between the foot of the tree and the point where the top touches the ground is 8 m, find the original height of the tree.

Answer:

Let the original height be H = h₁ + h₂, where h₁ is the unbroken part and h₂ is the broken part.
Using the cosine ratio: cos 30° = 8 / h₂ ⇒ h₂ = 8 / (√3/2) = 16/√3 m
Using the sine ratio: sin 30° = h₁ / h₂ ⇒ h₁ = (16/√3) × (1/2) = 8/√3 m
Thus, H = 8/√3 + 16/√3 = 24/√3 = 8√3 m
The original height was 8√3 meters.

Question 13:

A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 2 m away from the wall, find the length of the ladder.

Answer:

Let the length of the ladder be L.

Given: Base = 2 m, Angle = 60°
Using cosine in right triangle: cosθ = Base/Hypotenuse
cos60° = 2/L
1/2 = 2/L
L = 4 m

The ladder is 4 meters long.

Question 14:

The angle of elevation of the top of a tower from a point on the ground is 30°. If the point is 20 m away from the foot of the tower, find the height of the tower.

Answer:

Let the height of the tower be h.

Given: Distance = 20 m, Angle = 30°
Using tangent in right triangle: tanθ = Opposite/Adjacent
tan30° = h/20
1/√3 = h/20
h = 20/√3 m (Rationalize)
h = (20√3)/3 m

The tower's height is (20√3)/3 meters.

Question 15:

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. If the distance between the foot of the tree to the point where the top touches the ground is 6 m, find the original height of the tree.

Answer:

Let the original height be H = h₁ + h₂, where h₁ is the standing part and h₂ is the broken part.

Step 1: For standing part (cos30° = h₁/6) ⇒ h₁ = 6 × √3/2 = 3√3 m
Step 2: For broken part (sin30° = h₂/6) ⇒ h₂ = 6 × 1/2 = 3 m
Total height H = 3√3 + 3 = 3(√3 + 1) m

The original height was 3(√3 + 1) meters.

Question 16:

From the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. If the ships are on the same side of the lighthouse, find the distance between them.

Answer:

Given: Height of lighthouse = 75 m, angles of depression = 30° and 45°.
For the first ship (30°):
tan 30° = 75/distance1 ⇒ distance1 = 75√3 m
For the second ship (45°):
tan 45° = 75/distance2 ⇒ distance2 = 75 m
Distance between ships = distance1 - distance2 = 75(√3 - 1) m.

Question 17:

Answer:

Given: Angle with ground = 30°, distance = 6 m.
Let the height of the remaining part be h and the broken part be L.
Using cosine: cos 30° = 6/L ⇒ L = 6/(√3/2) = 4√3 m.
Using tangent: tan 30° = h/6 ⇒ h = 6/√3 = 2√3 m.
Original height = h + L = 6√3 m.

Question 18:

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

Answer:

Let the height of the tower be h.

Given: Distance = 30 m, Angle = 30°
Using tangent of the angle:
tan 30° = Height / Distance
1/√3 = h / 30
h = 30/√3 = 10√3 m

The height of the tower is 10√3 m.

Question 19:

A kite is flying at a height of 60 m with a string inclined at 45° to the horizontal. Find the length of the string.

Answer:

Let the length of the string be L.

Given: Height = 60 m, Angle = 45°
Using sine of the angle:
sin 45° = Height / Hypotenuse
1/√2 = 60 / L
L = 60√2 m

The length of the string is 60√2 m.

Question 20:

The shadow of a vertical pole is √3 times its height. Find the angle of elevation of the sun at that time.

Answer:

Let the height of the pole be h and the angle of elevation be θ.

Given: Shadow length = √3h
Using tangent of the angle:
tan θ = Height / Shadow
tan θ = h / (√3h) = 1/√3
θ = 30°

The angle of elevation of the sun is 30°.

Question 21:

From the top of a 10 m high building, the angle of depression of a point on the ground is 30°. Find the distance of the point from the foot of the building.

Answer:

Let the distance be d.

Given: Height = 10 m, Angle of depression = 30°
Angle of elevation = Angle of depression = 30°
Using tangent of the angle:
tan 30° = Height / Distance
1/√3 = 10 / d
d = 10√3 m

The distance is 10√3 m.

Question 22:

A tree breaks due to a storm and the broken part bends so that the top touches the ground at an angle of 30°. If the distance from the foot of the tree to the point where the top touches the ground is 5 m, find the original height of the tree.

Answer:

Let the original height be H and the broken part be h.

Given: Base = 5 m, Angle = 30°
Using tangent of the angle:
tan 30° = h / 5
h = 5/√3 m
Using cosine of the angle:
cos 30° = 5 / (H - h)
√3/2 = 5 / (H - 5/√3)
H = (10√3)/3 + 5/√3 = 15/√3 = 5√3 m

The original height of the tree was 5√3 m.

Question 23:

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.

Answer:

Let the height of the tower be H.

Step 1: Distance between buildings = 7 m (since angle of depression is 45°, height = distance).
Step 2: Using tangent for elevation: tan 60° = (H - 7)/7
√3 = (H - 7)/7
H - 7 = 7√3
H = 7 + 7√3 m

The tower's height is 7(1 + √3) meters.

Question 24:

A statue 1.6 m tall stands on the 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.

Answer:

Let the height of the pedestal be h.

Step 1: Distance from point to pedestal = h (since angle of elevation is 45°).
Step 2: For statue: tan 60° = (h + 1.6)/h
√3 = (h + 1.6)/h
√3h = h + 1.6
h(√3 - 1) = 1.6
h = 1.6/(√3 - 1) = 0.8(√3 + 1) m

The pedestal is 0.8(√3 + 1) meters tall.

Question 25:

The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower.

Answer:

Let the height of the tower be h.

Step 1: Initial shadow length = h (since angle is 45°).
Step 2: New shadow length = h + 10 m (angle becomes 30°).
Using tangent: tan 30° = h/(h + 10)
1/√3 = h/(h + 10)
h + 10 = h√3
h(√3 - 1) = 10
h = 10/(√3 - 1) = 5(√3 + 1) m

The tower's height is 5(√3 + 1) meters.

Question 26:

A flagstaff stands on the top of a building. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. If the height of the building is 12 m, find the height of the flagstaff.

Answer:

Given: Building height = 12 m, angle to bottom = 45°, angle to top = 60°.
Let the distance from the point to the building be d, and the flagstaff height be h.
Using angle to bottom (45°): tan(45°) = 12/d ⇒ 1 = 12/d ⇒ d = 12 m.
Using angle to top (60°): tan(60°) = (12 + h)/12 ⇒ √3 = (12 + h)/12 ⇒ 12 + h = 12√3 ⇒ h = 12√3 - 12 = 12(√3 - 1) m.
Answer: The flagstaff is 12(√3 - 1) m tall.

Question 27:

From the top of a 75 m high lighthouse, the angles of depression of two ships on the same side are 30° and 45°. Find the distance between the two ships.

Answer:

Given: Height (h) = 75 m, Angles of depression = 30° and 45°.
For the first ship (30°): tan(30°) = 75/d₁ → d₁ = 75√3 m.
For the second ship (45°): tan(45°) = 75/d₂ → d₂ = 75 m.
Distance between ships = d₁ - d₂ = 75√3 - 75 = 75(√3 - 1) m.
Answer: The distance is 75(√3 - 1) m.

Question 28:

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground at an angle of 30°. If the distance from the foot of the tree to the point where the top touches the ground is 6 m, find the original height of the tree.

Answer:

Given: Angle (θ) = 30°, Distance (base) = 6 m.
Let the height of the standing part be h and the broken part be L.
Using cosine: cos(30°) = 6/L → L = 6/cos(30°) = 6/(√3/2) = 12/√3 = 4√3 m.
Using tangent: tan(30°) = h/6 → h = 6 × (1/√3) = 2√3 m.
Total height = h + L = 2√3 + 4√3 = 6√3 m.
Answer: The original height was 6√3 m.

Question 29:

A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

Answer:

Let the length of the ladder be L.

Given: θ = 60°, distance from wall = 2.5 m

Using cosine of angle: cosθ = adjacent/hypotenuse
cos60° = 2.5/L
0.5 = 2.5/L
L = 2.5/0.5 = 5 m

The length of the ladder is 5 meters.

Question 30:

Answer:

Let the remaining height be h and broken part be L.

Given: distance = 8 m, θ = 30°

Using tangent: tan30° = h/8
1/√3 = h/8
h = 8/√3 m

Using cosine: cos30° = 8/L
√3/2 = 8/L
L = 16/√3 m

Total height = h + L = 8/√3 + 16/√3 = 24/√3 = 8√3 m

The original height of the tree was 8√3 meters.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

Answer:

Given: Angle of elevation = 60° and distance from the wall = 2.5 m.
Let the length of the ladder be L.

Using the cosine ratio in right-angled triangle:
cos 60° = adjacent side / hypotenuse
cos 60° = 2.5 / L

We know cos 60° = 0.5, so:
0.5 = 2.5 / L
L = 2.5 / 0.5
L = 5 m

Thus, the length of the ladder is 5 meters.

Question 2:

From the top of a 10 m high building, the angle of elevation of the top of a tower is 45° and the angle of depression of its foot is 30°. Find the height of the tower.

Answer:

Given: Height of building = 10 m, angle of elevation = 45°, and angle of depression = 30°.
Let the height of the tower be H and the distance between the building and the tower be d.

Step 1: Find d using the angle of depression (30°).
tan 30° = height of building / d
1/√3 = 10 / d
d = 10√3 m

Step 2: Find the additional height of the tower above the building using the angle of elevation (45°).
tan 45° = (H - 10) / d
1 = (H - 10) / 10√3
H - 10 = 10√3
H = 10 + 10√3 m

Thus, the height of the tower is 10 + 10√3 meters.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

A tower stands vertically on the ground. From a point on the ground 30 m away from the foot of the tower, the angle of elevation of the top is 30°. Find the height of the tower. (Use √3 = 1.732)

Answer:

Introduction

We studied that trigonometry helps find heights using angles. Here, we use the angle of elevation to solve.

Argument 1

Given: Distance from tower (BC) = 30 m, Angle (θ) = 30°.
Let height (AB) = h.

Argument 2

In ΔABC, tan 30° = AB/BC ⇒ 1/√3 = h/30 ⇒ h = 30/√3 = 10√3 ≈ 17.32 m.

Conclusion

Thus, the tower's height is 17.32 m, matching NCERT Example 1.

Question 2:

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.

Answer:

Introduction

Our textbook shows how trigonometry solves real-life problems like kite flying. Here, we find the string's length.

Argument 1

Given: Height (AB) = 60 m, Angle (θ) = 60°.
Let string length (AC) = L.

Argument 2

In ΔABC, sin 60° = AB/AC ⇒ √3/2 = 60/L ⇒ L = 120/√3 = 40√3 ≈ 69.28 m.

Conclusion

The string's length is 40√3 m, similar to NCERT Exercise 9.1 Q3.

Question 3:

The angle of elevation of the top of a building from a point on the ground is 45°. On walking 20 m towards the building, the angle becomes 60°. Find the building's height.

Answer:

Introduction

We use two angles of elevation to find heights. This is a common NCERT problem.

Argument 1

Let initial distance = x, height = h.
From first point: tan 45° = h/x ⇒ h = x.

Argument 2

From second point (20 m closer): tan 60° = h/(x-20) ⇒ √3 = h/(h-20) ⇒ h = 20√3/(√3-1) ≈ 47.32 m.

Conclusion

The building's height is 47.32 m, derived like NCERT Example 4.

Question 4:

A statue stands on a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and to the top of the pedestal is 45°. If the pedestal is 15 m high, find the statue's height.

Answer:

Introduction

Trigonometry helps separate heights in layered structures. Here, we find the statue's height.

Argument 1

Pedestal height (BC) = 15 m.
Let statue height (AB) = h, total height = (15 + h).

Argument 2

From ΔADC, tan 60° = (15+h)/DC. From ΔBDC, tan 45° = 15/DC ⇒ DC = 15 m. Thus, √3 = (15+h)/15 ⇒ h = 15√3 - 15 ≈ 10.98 m.

Conclusion

The statue's height is 10.98 m, solved like NCERT Exercise 9.1 Q7.

Question 5:

A tree breaks due to a storm and bends so that its top touches the ground at 30° and is 10 m from the base. Find the original height of the tree.

Answer:

Introduction

We studied broken objects using trigonometry. This problem combines Pythagoras and angles.

Argument 1

Let broken part (AC) = L, remaining part (BC) = h.
Given: Distance (CD) = 10 m, Angle (θ) = 30°.

Argument 2

In ΔADC, sin 30° = AD/AC ⇒ 1/2 = h/L ⇒ L = 2h. Also, cos 30° = 10/L ⇒ √3/2 = 10/(2h) ⇒ h = 10/√3 ≈ 5.77 m. Total height = h + L = 3h ≈ 17.32 m.

Conclusion

The tree's original height was 17.32 m, similar to NCERT Example 9.

Question 6:

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

Answer:

Introduction

We studied that trigonometry helps find heights using angles. Here, we use the angle of elevation to determine the tower's height.

Argument 1

Given: Distance from tower (BC) = 30 m, angle of elevation (θ) = 30°.

Argument 2

In ΔABC, tanθ = AB/BC ⇒ tan30° = AB/30 ⇒ AB = 30 × (1/√3) ≈ 17.32 m.

Conclusion

Thus, the tower's height is 17.32 m, derived using basic trigonometric ratios.

Question 7:

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. If the inclination of the string with the ground is 60°, find the length of the string.

Answer:

Introduction

Our textbook shows how trigonometry solves real-world problems like finding the length of a kite's string using height and angle.

Argument 1

Given: Height (AB) = 60 m, angle (θ) = 60°.

Argument 2

In ΔABC, sinθ = AB/AC ⇒ sin60° = 60/AC ⇒ AC = 60/(√3/2) ≈ 69.28 m.

Conclusion

The string's length is 69.28 m, calculated using sine ratio.

Question 8:

The angle of elevation of the top of a building from a point on the ground is 45°. On walking 20 m towards the building, the angle becomes 60°. Find the height of the building.

Answer:

Introduction

We use two angles of elevation to find the height of a building, a common NCERT application.

Argument 1

Let height = h, initial distance = x. From ΔABC, tan45° = h/x ⇒ h = x.

Argument 2

From ΔABD, tan60° = h/(x-20) ⇒ h = √3(x-20). Solving both: x = √3(x-20) ⇒ x ≈ 47.32 m ⇒ h ≈ 47.32 m.

Conclusion

The building's height is 47.32 m, derived using two trigonometric ratios.

Question 9:

A tree breaks due to a storm and bends so that its top touches the ground at 30° and is 8 m from its base. Find the original height of the tree.

Answer:

Introduction

Trigonometry helps reconstruct broken objects' heights. Here, we find the tree's original height using angle and distance.

Argument 1

Let the broken part = AC, standing part = AB. Given: BC = 8 m, angle (θ) = 30°.

Argument 2

In ΔABC, tan30° = AB/8 ⇒ AB ≈ 4.62 m. Also, cos30° = 8/AC ⇒ AC ≈ 9.24 m. Total height = AB + AC ≈ 13.86 m.

Conclusion

The tree's original height was 13.86 m, combining both parts.

Question 10:

Answer:

Introduction

We separate pedestal and statue heights using two angles of elevation, a textbook example.

Argument 1

Let statue height = h, total height = (15 + h). From ΔABC, tan45° = 15/BC ⇒ BC = 15 m.

Argument 2

From ΔABD, tan60° = (15+h)/15 ⇒ 15+h = 15√3 ⇒ h ≈ 10.98 m.

Conclusion

The statue's height is 10.98 m, calculated using trigonometric ratios.

Question 11:

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°. The car is approaching the foot of the tower with a uniform speed. After 6 seconds, the angle of depression changes to 60°. Find the time taken by the car to reach the foot of the tower from the initial point of observation.

Answer:

Let’s solve the problem step-by-step using trigonometric concepts:

Step 1: Draw the diagram

Draw a vertical line representing the tower (AB) and a horizontal line representing the highway. Mark the initial position of the car as point C and the position after 6 seconds as point D. The foot of the tower is point B.

Step 2: Assign variables

Let the height of the tower AB = h meters.
Let the distance of the car from the tower initially (BC) = x meters.
Let the speed of the car = v m/s.

Step 3: Use trigonometric ratios

In ΔABC (angle of depression = 30°):
tan 30° = AB / BC
1/√3 = h / x
=> h = x / √3 ...(1)

In ΔABD (angle of depression = 60°):
tan 60° = AB / BD
√3 = h / (x - 6v)
=> h = √3 (x - 6v) ...(2)

Step 4: Equate and solve

From (1) and (2):
x / √3 = √3 (x - 6v)
x = 3(x - 6v)
x = 3x - 18v
2x = 18v
x = 9v

Step 5: Find time taken

Total distance to cover = BC = x = 9v.
Time = Distance / Speed = 9v / v = 9 seconds.

Final Answer: The car will take 9 seconds to reach the foot of the tower from the initial point of observation.

Note: Always verify calculations and ensure units are consistent. Diagrams help visualize the problem clearly.

Question 12:

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 meters high, find the height of the building.

Answer:

Let the height of the building be h meters and the distance between the tower and the building be d meters.

From the foot of the tower:

tan(60°) = 50 / d
=> √3 = 50 / d
=> d = 50 / √3 meters.

From the foot of the building:

tan(30°) = h / d
=> 1/√3 = h / (50/√3)
=> h = (50/√3) × (1/√3)
=> h = 50 / 3 meters.

Thus, the height of the building is 50/3 meters or approximately 16.67 meters.

Question 13:

Answer:

Let the height of the tower be h meters and the speed of the car be v m/s.

Let the initial distance of the car from the tower be x meters.

From the first observation (angle of depression = 30°):

tan(30°) = h / x

=> 1/√3 = h / x

=> x = h√3 ...(1)

After 6 seconds, the car covers a distance of 6v meters, so the new distance from the tower is x - 6v meters.

From the second observation (angle of depression = 60°):

tan(60°) = h / (x - 6v)

=> √3 = h / (x - 6v)

=> x - 6v = h / √3 ...(2)

Substitute equation (1) into equation (2):

h√3 - 6v = h / √3

Multiply both sides by √3:

3h - 6v√3 = h

=> 2h = 6v√3

=> h = 3v√3

Now, the remaining distance for the car to reach the tower is x - 6v = h / √3 = (3v√3) / √3 = 3v meters.

Time taken to cover this distance = 3v / v = 3 seconds.

Thus, the car will take 3 seconds to reach the foot of the tower from the second observation point.

Question 14:

Answer:

Let the height of the building be h meters and the distance between the tower and the building be x meters.

From the foot of the tower:

In ΔABC (angle of elevation to the top of the building is 30°):

tan(30°) = h / x
=> 1/√3 = h / x
=> x = h√3 ...(1)

From the foot of the building:

In ΔABD (angle of elevation to the top of the tower is 60°):

tan(60°) = 50 / x
=> √3 = 50 / x
=> x = 50 / √3 ...(2)

Equating (1) and (2):

h√3 = 50 / √3
=> h = (50 / √3) × (1/√3)
=> h = 50 / 3
=> h ≈ 16.67 m.

Thus, the height of the building is approximately 16.67 meters.

Question 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°. The car is approaching the foot of the tower with a uniform speed. After 6 seconds, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let the height of the tower be h meters and the initial distance of the car from the tower be x meters.

In ΔABC (when angle of depression is 30°):

tan(30°) = h / x
=> 1/√3 = h / x
=> x = h√3 ...(1)

After 6 seconds, the car moves closer to the tower. Let the new distance be y meters.

In ΔABD (when angle of depression is 60°):

tan(60°) = h / y
=> √3 = h / y
=> y = h / √3 ...(2)

The distance covered by the car in 6 seconds = x - y
= h√3 - h/√3
= (3h - h)/√3
= 2h/√3 meters.

Speed of the car = Distance / Time = (2h/√3) / 6 = h/(3√3) m/s.

Time taken to cover remaining distance y = Distance / Speed
= (h/√3) / (h/(3√3))
= 3 seconds.

Thus, the car will take 3 seconds to reach the foot of the tower from the second observation point.

Question 16:

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 it. The distance between the foot of the tree to the point where the top touches the ground is 8 meters. Find the original height of the tree before it broke.

Answer:

To solve this problem, we will use the concept of trigonometry and the properties of a right-angled triangle.

Given:

The angle between the broken part of the tree and the ground is 30°.
The distance from the foot of the tree to the point where the top touches the ground is 8 meters.

Step 1: Draw the diagram

Visualize the scenario as a right-angled triangle where:

The broken part of the tree forms the hypotenuse.
The distance on the ground (8 meters) is the adjacent side to the angle of 30°.
The remaining part of the tree is the opposite side.

Step 2: Find the height of the broken part

Let the height of the broken part be h.

Using the cosine of the angle:

cos(30°) = adjacent / hypotenuse

cos(30°) = 8 / h

h = 8 / cos(30°)

h = 8 / (√3/2)

h = (8 × 2) / √3

h = 16 / √3

Rationalizing the denominator:

h = (16√3) / 3 meters

Step 3: Find the remaining height of the tree

Let the remaining height be r.

Using the tangent of the angle:

tan(30°) = opposite / adjacent

tan(30°) = r / 8

r = 8 × tan(30°)

r = 8 × (1/√3)

r = 8 / √3

Rationalizing the denominator:

r = (8√3) / 3 meters

Step 4: Calculate the original height

The original height of the tree is the sum of the broken part and the remaining part:

Original height = h + r

Original height = (16√3)/3 + (8√3)/3

Original height = (24√3)/3

Original height = 8√3 meters

Final Answer: The original height of the tree was 8√3 meters.

Value-added information:

Trigonometry is widely used in real-life scenarios like measuring heights, distances, and angles. Understanding these concepts helps in fields such as engineering, architecture, and navigation.

Question 17:

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°. After the car approaches the tower by 50 meters, the angle of depression becomes 45°. Find the height of the tower and the initial distance of the car from the tower. (Use √3 = 1.732)

Answer:

Let the height of the tower be h meters and the initial distance of the car from the tower be x meters.

From the first observation (angle of depression = 30°):

tan(30°) = h / x
=> 1/√3 = h / x
=> h = x / √3 ...(1)

After the car moves 50 meters closer, the new distance is (x - 50) meters.

From the second observation (angle of depression = 45°):

tan(45°) = h / (x - 50)
=> 1 = h / (x - 50)
=> h = x - 50 ...(2)

Equating (1) and (2):

x / √3 = x - 50
=> x = √3 (x - 50)
=> x = √3x - 50√3
=> √3x - x = 50√3
=> x (√3 - 1) = 50√3
=> x = 50√3 / (√3 - 1)

Rationalizing the denominator:

x = 50√3 (√3 + 1) / ( (√3 - 1)(√3 + 1) )
=> x = 50 (3 + √3) / (3 - 1)
=> x = 50 (3 + √3) / 2
=> x = 25 (3 + √3)
=> x = 25 (3 + 1.732) = 25 × 4.732 = 118.3 meters (approx)

Now, substituting x in equation (2):

h = x - 50
=> h = 118.3 - 50 = 68.3 meters (approx)

Thus, the height of the tower is 68.3 meters and the initial distance of the car from the tower was 118.3 meters.

Question 18:

Answer:

Let the height of the tower be h meters, and the initial distance of the car from the tower be x meters.

From the first observation (angle of depression = 30°):

tan(30°) = h/x ⇒ 1/√3 = h/x ⇒ x = h√3 ...(1)

After 6 seconds, the angle of depression becomes 60°. Let the new distance be y meters.

tan(60°) = h/y ⇒ √3 = h/y ⇒ y = h/√3 ...(2)

The distance covered by the car in 6 seconds = x - y = h√3 - h/√3 = (3h - h)/√3 = 2h/√3 meters.

Speed of the car = Distance/Time = (2h/√3)/6 = h/(3√3) m/s.

Time taken to cover the remaining distance y = h/√3 meters:

Time = Distance/Speed = (h/√3) / (h/(3√3)) = 3 seconds.

Thus, the total time taken by the car to reach the foot of the tower from the initial observation point is 6 + 3 = 9 seconds.

Question 19:

A statue stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 45° and from the same point, the angle of elevation of the top of the pedestal is 30°. If the height of the pedestal is 10 m, find the height of the statue.

Answer:

Let the height of the statue be h meters and the distance of the observation point from the pedestal be x meters.

For the pedestal (height = 10 m):

tan(30°) = 10 / x
=> 1/√3 = 10 / x
=> x = 10√3 m ...(1)

For the statue (total height = 10 + h):

tan(45°) = (10 + h) / x
=> 1 = (10 + h) / x
=> x = 10 + h ...(2)

From (1) and (2):

10√3 = 10 + h
=> h = 10√3 - 10
=> h = 10(√3 - 1) m

Thus, the height of the statue is 10(√3 - 1) meters (approximately 7.32 m).

Note: The exact form is preferred in CBSE answers unless specified otherwise.

Question 20:

From the top of a 75 m high lighthouse, the angles of depression of two ships on the same side of the lighthouse are observed as 30° and 45°. If the ships are in line with the foot of the lighthouse, find the distance between the two ships.

Answer:

Let the distance of the first ship (with angle of depression 30°) from the lighthouse be x meters, and the distance of the second ship (with angle of depression 45°) be y meters.

Height of the lighthouse = 75 m.

For the first ship:

tan(30°) = 75/x ⇒ 1/√3 = 75/x ⇒ x = 75√3 ...(1)

For the second ship:

tan(45°) = 75/y ⇒ 1 = 75/y ⇒ y = 75 ...(2)

The distance between the two ships = x - y = 75√3 - 75 = 75(√3 - 1) meters.

Thus, the distance between the two ships is 75(√3 - 1) meters.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2m away from the wall. Find the length of the ladder and the height it reaches on the wall.

Answer:

Problem Interpretation

We studied that a ladder forms a right triangle with the wall and ground. Here, the angle is 60° and base distance is 2m.

Mathematical Modeling

Let ladder length = hypotenuse (h)
Height on wall = perpendicular (p)

Solution

Using cos60° = base/hypotenuse: 0.5 = 2/h ⇒ h = 4m. Then, sin60° = p/4 ⇒ p = 4×√3/2 = 2√3m. Final answer: Ladder = 4m, Height = 2√3m.

Question 2:

From a point 100m above a lake, the angle of depression of a cloud is 30°. The reflection of the cloud in the lake has an angle of depression of 60°. Find the height of the cloud above the lake.

Answer:

Problem Interpretation

Our textbook shows similar problems with angles of depression. We need to find cloud height (H) using two observations.

Mathematical Modeling

Let observer height = 100m
Cloud height = H

Solution

Using trigonometry: tan30° = (H-100)/x and tan60° = (H+100)/x. Solving these, we get H = 200m. Final answer: Cloud height = 200m.

Question 3:

A tree breaks due to storm and the broken part bends touching the ground 30° angle. The distance between foot of tree and point where top touches is 8m. Find original height of the tree.

Answer:

Problem Interpretation

This is a right triangle problem where the broken tree forms two sides: standing part (h) and fallen part (l).

Mathematical Modeling

Let standing height = h
Fallen part = hypotenuse = l

Solution

Using tan30° = h/8 ⇒ h = 8/√3m. Then, cos30° = 8/l ⇒ l = 16/√3m. Total height = h + l = 24/√3 = 8√3m. Final answer: 8√3m.

Question 4:

A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2 m away from the wall.
Problem Interpretation: Find the length of the ladder and the height it reaches on the wall.

Answer:

Problem Interpretation: We need to find the ladder's length (hypotenuse) and wall height (opposite side) using trigonometry.
Mathematical Modeling: Let ladder length be L and height be h. Given: base = 2 m, angle = 60°.
Solution:

cos 60° = base/hypotenuse ⇒ 0.5 = 2/L ⇒ L = 4 m
sin 60° = height/hypotenuse ⇒ √3/2 = h/4 ⇒ h = 2√3 m

Question 5:

From a point 100 m above a lake, the angle of depression of a cloud is 30°. The reflection of the cloud in the lake has an angle of depression of 60°.
Problem Interpretation: Find the height of the cloud above the lake.

Answer:

Problem Interpretation: We must find the cloud's height (H) using angle of depression concepts.
Mathematical Modeling: Let horizontal distance be x. From the observer: tan 30° = (H-100)/x. For reflection: tan 60° = (H+100)/x.
Solution:

Divide equations: √3 = (H+100)/(H-100)
Solve: H = 200 m

Question 6:

A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2 m away from the wall. Find the length of the ladder and the height it reaches on the wall. [Diagram: Right triangle formed by ladder, wall, and ground]

Answer:

Problem Interpretation

We studied that a ladder forms a right triangle with the wall and ground. Here, the angle is 60° and base distance is 2 m.

Mathematical Modeling

Let ladder length = hypotenuse (h)
Height on wall = perpendicular (p)

Solution

Using cos 60° = base/hypotenuse: 0.5 = 2/h → h = 4 m. Then, sin 60° = p/4 → p = 4×√3/2 = 2√3 m.

Question 7:

From a point 100 m above a lake, the angle of depression of a cloud is 30°. The reflection of the cloud in the lake has an angle of depression of 60°. Find the height of the cloud above the lake. [Diagram: Observer, cloud, and its reflection forming two right triangles]

Answer:

Problem Interpretation

Our textbook shows that angle of depression equals angle of elevation. Here, we have two triangles with angles 30° and 60°.

Mathematical Modeling

Let cloud height = H
Horizontal distance = x

Solution

tan 30° = (H-100)/x → x = (H-100)√3. For reflection: tan 60° = (H+100)/x → x = (H+100)/√3. Solving both: H = 200 m.

Question 8:

Answer:

Problem Interpretation

We studied that a ladder forms a right triangle with the wall and ground. Here, the angle is 60° and base distance is 2 m.

Mathematical Modeling

Let ladder length = L
Height on wall = h

Solution

Using cos60° = base/hypotenuse: 0.5 = 2/L ⇒ L = 4 m. Using Pythagoras theorem: h = √(4² - 2²) = √12 = 3.464 m.

Question 9:

From a point 100 m above a lake, the angle of depression of a cloud is 30°. The reflection of the cloud in the lake has an angle of depression of 60°. Find the height of the cloud above the lake.

Answer:

Problem Interpretation

Our textbook shows similar problems with angles of depression. We need to find the cloud's height (H) using trigonometry.

Mathematical Modeling

Let horizontal distance = x
Observer height = 100 m

Solution

tan30° = (H-100)/x and tan60° = (H+100)/x. Solving these: x = √3(H-100) and x = (H+100)/√3. Equating gives H = 200 m.

Question 10:

A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2 m away from the wall. Find the length of the ladder and the height it reaches on the wall.

Answer:

Problem Interpretation

We studied that a ladder forms a right triangle with the wall and ground. Here, the angle is 60° and base distance is 2 m.

Mathematical Modeling

Let ladder length = hypotenuse (L)
Height on wall = perpendicular (h)

Solution

Using cos 60° = base/hypotenuse, we get 0.5 = 2/L ⇒ L = 4 m. Then, sin 60° = h/4 ⇒ h = 4 × √3/2 = 2√3 m.

Question 11:

From a point 100 m above a lake, the angle of depression of a cloud is 30°. The reflection of the cloud in the lake has an angle of depression of 60°. Calculate the height of the cloud above the lake.

Answer:

Problem Interpretation

Our textbook shows that angle of depression equals angle of elevation. We have two right triangles here.

Mathematical Modeling

Let cloud height = H
Horizontal distance = d

Solution

tan 30° = (H−100)/d and tan 60° = (H+100)/d. Solving, we get √3(H−100) = (H+100)/√3 ⇒ 3H−300 = H+100 ⇒ H = 200 m.

Question 12:

A lighthouse is situated at the top of a cliff near a seashore. From a boat in the sea, the angle of elevation of the top of the lighthouse is 30°, and the angle of elevation of the top of the cliff is 15°. The height of the lighthouse is 50 meters. Draw a diagram representing the scenario and find the distance of the boat from the base of the cliff.

Answer:

Let's solve the problem step-by-step:

Step 1: Draw the Diagram
Draw a right-angled triangle where:

The cliff is the vertical side (height = h).
The lighthouse is an extension on top of the cliff (total height = h + 50m).
The boat is at a horizontal distance 'd' from the cliff base.

Step 2: Use Trigonometry for the Cliff
For the cliff (angle of elevation = 15°):
tan(15°) = h / d
=> h = d × tan(15°) ...(1)

Step 3: Use Trigonometry for the Lighthouse
For the lighthouse (angle of elevation = 30°):
tan(30°) = (h + 50) / d
=> h + 50 = d × tan(30°) ...(2)

Step 4: Substitute Equation (1) into (2)
d × tan(15°) + 50 = d × tan(30°)
=> 50 = d (tan(30°) - tan(15°))
=> d = 50 / (tan(30°) - tan(15°))

Step 5: Calculate Numerical Value
Using tan(30°) ≈ 0.577 and tan(15°) ≈ 0.268:
d = 50 / (0.577 - 0.268) ≈ 50 / 0.309 ≈ 161.8 meters.

Thus, the boat is approximately 161.8 meters away from the base of the cliff.

Question 13:

A tree breaks due to a storm, and its top touches the ground at an angle of 30°, forming a right triangle with the ground. The distance from the base of the tree to the point where the top touches the ground is 10 meters. Find the original height of the tree before it broke.

Answer:

Let's solve the problem step-by-step:

Step 1: Draw the Diagram
Draw a right-angled triangle where:

The broken part of the tree is the hypotenuse.
The remaining vertical part is one side (height = h).
The horizontal distance (10m) is the other side.

Step 2: Use Trigonometry
Given angle = 30° and adjacent side = 10m:
tan(30°) = opposite side / adjacent side
=> tan(30°) = h / 10
=> h = 10 × tan(30°)
=> h ≈ 10 × 0.577 ≈ 5.77 meters.

Step 3: Find the Hypotenuse (Broken Part)
cos(30°) = adjacent side / hypotenuse
=> hypotenuse = 10 / cos(30°)
=> hypotenuse ≈ 10 / 0.866 ≈ 11.55 meters.

Step 4: Calculate Original Height
Original height = h + hypotenuse
=> Original height ≈ 5.77 + 11.55 ≈ 17.32 meters.

Thus, the original height of the tree was approximately 17.32 meters.

Question 14:

A lighthouse is situated at the top of a cliff. From a boat in the sea, the angle of elevation of the top of the lighthouse is 30°, and the angle of elevation of the top of the cliff is 15°. The boat is 100 meters away from the base of the cliff.

Calculate the height of the lighthouse above the cliff.

Answer:

Let the height of the cliff be h meters and the height of the lighthouse above the cliff be x meters.

The distance of the boat from the cliff is 100 meters.

For the cliff:
tan(15°) = h / 100
h = 100 × tan(15°)
h ≈ 100 × 0.2679 ≈ 26.79 meters

For the lighthouse:
tan(30°) = (h + x) / 100
h + x = 100 × tan(30°)
26.79 + x ≈ 100 × 0.5774 ≈ 57.74 meters
x ≈ 57.74 - 26.79 ≈ 30.95 meters

Thus, the height of the lighthouse above the cliff is approximately 30.95 meters.

Question 15:

A tree is broken by the wind such that its top touches the ground at a distance of 10 meters from its base, making an angle of 45° with the ground.

Find the original height of the tree before it broke.

Answer:

Let the height of the tree before breaking be H meters, and the height of the remaining part after breaking be h meters.

The broken part forms the hypotenuse of a right-angled triangle with the ground.

Given:
Distance from the base to the broken top = 10 meters
Angle with the ground = 45°

Using trigonometry:
tan(45°) = h / 10
1 = h / 10
h = 10 meters

The broken part of the tree is the hypotenuse:
sin(45°) = h / broken part length
Broken part length = h / sin(45°) ≈ 10 / 0.7071 ≈ 14.14 meters

Original height of the tree (H) = h + broken part length
H ≈ 10 + 14.14 ≈ 24.14 meters

Thus, the original height of the tree was approximately 24.14 meters.

Question 16:

A lighthouse is situated at the top of a cliff near a seashore. From a boat in the sea, the angle of elevation of the top of the lighthouse is 30°, and the angle of elevation of the top of the cliff is 15°. If the height of the lighthouse is 50 meters, find the height of the cliff. (Use √3 = 1.732)

Answer:

Let the height of the cliff be h meters and the horizontal distance between the boat and the cliff be x meters.

From the boat, the angle of elevation to the top of the lighthouse is 30°, and the total height observed is h + 50 meters.
Using trigonometry, we can write:
tan 30° = (h + 50) / x
1/√3 = (h + 50) / x
x = (h + 50)√3

Similarly, for the angle of elevation to the top of the cliff (15°):
tan 15° = h / x
From trigonometric tables, tan 15° = 2 - √3
So, 2 - √3 = h / x
x = h / (2 - √3)

Equating the two expressions for x:
(h + 50)√3 = h / (2 - √3)
Cross-multiplying:
(h + 50)√3 (2 - √3) = h
Expanding:
2h√3 - 3h + 100√3 - 150 = h
Bringing all terms to one side:
2h√3 - 4h + 100√3 - 150 = 0
Solving for h:
h(2√3 - 4) = 150 - 100√3
h = (150 - 100√3) / (2√3 - 4)
Rationalizing the denominator:
h = [(150 - 100√3)(2√3 + 4)] / [(2√3 - 4)(2√3 + 4)]
Simplifying denominator:
(2√3)² - (4)² = 12 - 16 = -4
Simplifying numerator:
300√3 + 600 - 600 - 400√3 = -100√3
Thus:
h = -100√3 / -4 = 25√3 ≈ 25 × 1.732 ≈ 43.3 meters (approx).

Therefore, the height of the cliff is approximately 43.3 meters.

Question 17:

Two towers, A and B, are standing on level ground. The angle of elevation of the top of tower A from the foot of tower B is 60°, and the angle of elevation of the top of tower B from the foot of tower A is 30°. If the height of tower B is 30 meters, find the height of tower A and the distance between the two towers.

Answer:

Let the height of tower A be h meters and the distance between the two towers be x meters.

From the foot of tower B, the angle of elevation to the top of tower A is 60°:
tan 60° = h / x
√3 = h / x
h = x√3

From the foot of tower A, the angle of elevation to the top of tower B is 30°:
tan 30° = 30 / x
1/√3 = 30 / x
x = 30√3

Substituting x into the equation for h:
h = (30√3) × √3 = 30 × 3 = 90 meters.

Thus, the height of tower A is 90 meters, and the distance between the two towers is 30√3 meters (approximately 51.96 meters).

Question 18:

A kite is flying at a height of 60 meters from 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 there is no slack in it. Also, determine the horizontal distance of the kite from the tied point.

Answer:

We'll solve this using right-angled triangle trigonometry.

Step 1: Find the length of the string (L).
Given the height (h) = 60 meters and angle (θ) = 60°:
sin(60°) = h / L
L = h / sin(60°)
L = 60 / (√3/2) ≈ 60 / 0.866 ≈ 69.28 meters

Step 2: Find the horizontal distance (D).
Using the cosine ratio:
cos(60°) = D / L
D = L × cos(60°)
D ≈ 69.28 × 0.5 ≈ 34.64 meters

The length of the string is approximately 69.28 meters, and the horizontal distance is 34.64 meters.

Question 19:

A tree is broken by the wind such that its top touches the ground at a distance of 10 meters from the base of the tree, making an angle of 45° with the ground.

Find the original height of the tree before it was broken.

Answer:

This problem involves applying the Pythagorean theorem and trigonometric ratios.

Step 1: Let the height of the remaining part of the tree (still standing) be h meters.
Step 2: The broken part forms the hypotenuse of a right-angled triangle with the ground.
Since the angle between the broken part and the ground is 45°, the triangle is isosceles.
Thus, the height of the broken part above the ground is equal to the distance from the base (10 meters).
Step 3: Using the Pythagorean theorem for the broken part:
Length of broken part = √(10² + 10²) = √200 ≈ 14.14 meters
Step 4: Total original height of the tree = height of standing part + height of broken part
= h + 14.14
But since the standing part is also 10 meters (as the triangle is isosceles),
Total height ≈ 10 + 14.14 ≈ 24.14 meters

Therefore, the original height of the tree was approximately 24.14 meters.

Some Applications of Trigonometry – CBSE NCERT Study Resources

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