Some Applications of Trigonometry — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Some Applications of Trigonometry" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Some Applications of Trigonometry" — 8 important questions with detailed answers f…
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NCERT Solutions for Class 10 Mathematics — Some Applications of Trigonometry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Some Applications of Trigonometry — All 8 Questions Solved
Q1. Define angle of elevation and angle of depression. Draw a diagram showing both concepts.
Angle of Elevation:
The angle of elevation is the angle between the horizontal line of sight and the line of sight upward to an object that is above the horizontal.
When an observer looks upward at an object, the angle formed below the object's line of sight from the horizontal is called the angle of elevation.
Angle of Depression:
The angle of depression is the angle between the horizontal line of sight and the line of sight downward to an object that is below the horizontal.
When an observer looks downward at an object, the angle formed above the object's line of sight from the horizontal is called the angle of depression.
Diagram Description (Heights and Distances):
Object B (above observer)
|
| } angle of elevation from A
| /
Observer A ---------- Horizontal line
| \ } angle of depression to C
| \
C (below observer)
Key Property:
If an observer at point A looks at point B (above) with angle of elevation α, then from B's perspective, the angle of depression to A is also α. These angles are equal because they are alternate interior angles formed by parallel lines (the two horizontal lines at A and B) and a transversal (the line AB).
Applications:
- Heights of buildings, trees, mountains
- Distances of objects from an observer
- Navigation and surveying
- Astronomy
Q2. A ladder 10 m long leans against a wall. The ladder makes an angle of 60° with the ground. Find the height of the wall reached by the ladder and the distance of the foot of the ladder from the wall.
Given:
Length of ladder = 10 m (hypotenuse)
Angle with ground = 60°
Find:
1. Height of wall reached by ladder (opposite side)
2. Distance from wall to foot of ladder (adjacent side)
Diagram (text description):
Wall
|
| Height (h)
|/
Ladder (10 m)
/|
/ |
/ |
/60°|
Ground
Distance (d)
Step 1: Find the height (opposite side).
sin 60° = height / ladder
sin 60° = h / 10
√3/2 = h / 10
h = 10 × (√3/2)
h = 5√3 m
h ≈ 5 × 1.732 = 8.66 m
Step 2: Find the distance from wall (adjacent side).
cos 60° = distance / ladder
cos 60° = d / 10
1/2 = d / 10
d = 10 × (1/2)
d = 5 m
Answer:
Height of wall = 5√3 m ≈ 8.66 m
Distance from wall = 5 m
Verification:
Using Pythagorean theorem:
(5√3)² + 5² = 75 + 25 = 100 = 10² ✓
Q3. A man standing on a level ground observes the angle of elevation to the top of a building as 30°. He walks 50 m closer to the building and observes the angle of elevation as 60°. Find the height of the building.
Given:
Initial angle of elevation = 30°
Angle of elevation after walking 50 m closer = 60°
Distance walked = 50 m
Find: Height of building
Diagram (text description):
Building (height = h)
|
|
| 60°
|/
---*--- (closer position, 50m away)
/ |
30° | h
/ |
*-------+-------* (original position)
50m d
Total distance = d + 50 m
Step 1: From the original position (distance = d + 50):
tan 30° = h / (d + 50)
1/√3 = h / (d + 50)
h = (d + 50) / √3 ... (Equation 1)
Step 2: From the closer position (distance = d):
tan 60° = h / d
√3 = h / d
h = d√3 ... (Equation 2)
Step 3: Substitute Equation 2 into Equation 1.
d√3 = (d + 50) / √3
d√3 × √3 = d + 50
3d = d + 50
3d - d = 50
2d = 50
d = 25 m
Step 4: Find the height using Equation 2.
h = d√3
h = 25√3 m
h ≈ 25 × 1.732 = 43.3 m
Answer: Height of building = 25√3 m ≈ 43.3 m
Verification:
From position 1 (distance = 75 m):
tan 30° = 25√3 / 75 = √3 / 3 = 1/√3 ✓
From position 2 (distance = 25 m):
tan 60° = 25√3 / 25 = √3 ✓
Q4. An observer on a lighthouse 60 m high observes two ships at angles of depression 30° and 60°. The ships are on the same side of the lighthouse and are on the same horizontal line. Find the distance between the ships.
Given:
Height of lighthouse = 60 m
Angle of depression to ship 1 = 30°
Angle of depression to ship 2 = 60°
Both ships are on the same side and same horizontal line
Find: Distance between the ships
Diagram (text description):
Lighthouse
|\60 m
| \
| \ 30°
| \
| \_____ Ship 1 (distance d₁ from base)
|
| \ 60°
| \
|___\
Base| Ship 2 (distance d₂ from base)
Note: Angle of depression equals angle of elevation from the ship's perspective.
Step 1: Find distance d₂ to ship 2 (angle of depression 60°).
tan 60° = height / d₂
√3 = 60 / d₂
d₂ = 60 / √3
d₂ = 60 / √3 × √3/√3
d₂ = 60√3 / 3
d₂ = 20√3 m
Step 2: Find distance d₁ to ship 1 (angle of depression 30°).
tan 30° = height / d₁
1/√3 = 60 / d₁
d₁ = 60√3 m
Step 3: Find the distance between the ships.
Distance = d₁ - d₂ (since both are on the same side)
Distance = 60√3 - 20√3
Distance = 40√3 m
Distance ≈ 40 × 1.732 = 69.28 m
Answer: Distance between ships = 40√3 m ≈ 69.28 m
Verification:
tan 60° = 60 / 20√3 = 3 / √3 = √3 ✓
tan 30° = 60 / 60√3 = 1 / √3 ✓
Q5. From a point on the ground 100 m from the foot of a building, the angle of elevation to the top is 45°. Find the height of the building. If the angle changes to 60° after moving 25 m closer, verify your answer.
Given:
Initial distance from building = 100 m
Initial angle of elevation = 45°
After moving 25 m closer, new angle = 60°
Find: Height of building
Part 1: Using initial observation
Step 1: Use angle of elevation formula.
tan 45° = height / distance
1 = h / 100
h = 100 m
Part 2: Verification with second observation
Step 2: New distance after moving 25 m closer.
New distance = 100 - 25 = 75 m
Step 3: Check with the second angle.
tan 60° = h / new distance
√3 = h / 75
h = 75√3 m
h ≈ 75 × 1.732 = 129.9 m
Analysis:
The two methods give different heights:
Method 1: h = 100 m
Method 2: h = 75√3 ≈ 129.9 m
This suggests there's a discrepancy in the problem statement. However, if we assume the first observation is correct:
Answer: Height of building = 100 m
Verification of consistency:
If h = 100 m and initial distance = 100 m:
tan 45° = 100/100 = 1 ✓ (since tan 45° = 1)
If the observer moves 25 m closer (new distance = 75 m):
New angle = arctan(100/75) = arctan(4/3) ≈ 53.13°
This is close to 60° but not exactly 60°, indicating the second measurement has a slight discrepancy from the first.
Q6. A vertical pole of height 20 m casts a shadow on the ground. The angle of elevation of the sun is 45°. Find the length of the shadow.
Given:
Height of pole = 20 m (vertical)
Angle of elevation of sun = 45°
Find: Length of shadow
Diagram (text description):
Sun
/|
/ |
/ | 20 m (height of pole)
/ 45°|
/_____|_________
shadow (s)
Step 1: Apply tangent ratio.
tan(angle) = opposite / adjacent
tan 45° = height / shadow
tan 45° = 20 / s
Step 2: Use the value tan 45° = 1.
1 = 20 / s
s = 20 m
Answer: Length of shadow = 20 m
Note: When angle of elevation = 45°, the height equals the base (shadow length), forming an isosceles right triangle.
Verification:
tan 45° = 20/20 = 1 ✓
Angle of elevation = arctan(20/20) = arctan(1) = 45° ✓
Q7. An aeroplane flying horizontally at a height of 2000 m is observed from a point on the ground. The angle of elevation at a certain instant is 30°. After 10 seconds, the angle becomes 60°. Assuming the aeroplane flies in a straight line, find the speed of the aeroplane.
Given:
Height of aeroplane = 2000 m (constant)
Angle of elevation initially = 30°
Angle of elevation after 10 seconds = 60°
Find: Speed of aeroplane
Diagram (text description):
Aeroplane path →
Position 1 Position 2
✈ ✈ (2000 m altitude)
| |
| |
30°| |60°
|_____________|
Observer on ground
Step 1: Find the horizontal distance from observer to position 1.
tan 30° = height / d₁
1/√3 = 2000 / d₁
d₁ = 2000√3 m
Step 2: Find the horizontal distance from observer to position 2.
tan 60° = height / d₂
√3 = 2000 / d₂
d₂ = 2000 / √3
d₂ = 2000√3 / 3 m
Step 3: Find the distance traveled by aeroplane.
Distance = d₁ - d₂ (aeroplane moves towards observer)
Distance = 2000√3 - 2000√3/3
Distance = 2000√3(1 - 1/3)
Distance = 2000√3 × 2/3
Distance = 4000√3 / 3 m
Step 4: Find the speed.
Speed = Distance / Time
Speed = (4000√3/3) / 10
Speed = 400√3 / 3 m/s
Speed ≈ 400 × 1.732 / 3 = 692.8 / 3 ≈ 230.93 m/s
Step 5: Convert to km/h (optional).
Speed in km/h = 230.93 × 3.6 ≈ 831.35 km/h
Answer:
Speed of aeroplane = 400√3/3 m/s ≈ 231 m/s
Or approximately 831 km/h
Q8. Two poles of heights 10 m and 15 m are 20 m apart. Find the angle of elevation from the top of the shorter pole to the top of the taller pole.
Given:
Height of pole 1 = 10 m
Height of pole 2 = 15 m
Horizontal distance between poles = 20 m
Find: Angle of elevation from top of pole 1 to top of pole 2
Diagram (text description):
Pole 2 (15 m)
|
|_____ θ (angle of elevation from pole 1)
| /
| / 5 m (height difference)
Pole 1| /
(10 m) | /
___|/
|___________|
20 m
Step 1: Find the vertical height difference.
Height difference = 15 - 10 = 5 m
Step 2: The horizontal distance remains 20 m.
Step 3: Apply the tangent ratio.
tan θ = opposite / adjacent
tan θ = height difference / horizontal distance
tan θ = 5 / 20
tan θ = 1/4
tan θ = 0.25
Step 4: Find the angle.
θ = arctan(0.25)
θ ≈ 14.04°
Alternative: Using degrees and minutes
θ ≈ 14°02'24"
Answer: Angle of elevation = arctan(1/4) ≈ 14.04° or approximately 14°
Some Applications of Trigonometry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The line drawn from the eye of an observer to the object viewed is called the:
- A. Horizontal line
- B. Vertical line
- C. Line of sight
- D. Perpendicular line
Answer: C. Line of sight — The line of sight is the direct line from the observer's eye to the object being viewed.
MCQ 2. If a person is looking up at the top of a tower, the angle formed by the line of sight with the horizontal is called the:
- A. Angle of depression
- B. Angle of elevation
- C. Right angle
- D. Straight angle
Answer: B. Angle of elevation — The angle of elevation is the angle formed between the horizontal line and the line of sight when the object is above the horizontal level.
MCQ 3. When an observer looks down at an object, the angle formed by the line of sight with the horizontal is known as the:
- A. Angle of elevation
- B. Angle of incidence
- C. Angle of depression
- D. Angle of refraction
Answer: C. Angle of depression — The angle of depression is the angle formed between the horizontal line and the line of sight when the object is below the horizontal level.
MCQ 4. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. The height of the tower is:
- A. 15√3 m
- B. 15/√3 m
- C. 15 m
- D. 30 m
Answer: A. 15√3 m — Let H be the height of the tower. In the right-angled triangle, tan 60° = H / 15. Since tan 60° = √3, H = 15√3 m.
MCQ 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°. The length of the string, assuming no slack in the string, is:
- A. 60 m
- B. 120 m
- C. 40√3 m
- D. 60√3 m
Answer: C. 40√3 m — Let L be the length of the string. In the right-angled triangle, sin 60° = 60 / L. Since sin 60° = √3/2, L = 60 / (√3/2) = 120/√3 = 40√3 m.
MCQ 6. 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. If the angle made by the rope with the ground level is 30°, the height of the pole is:
- A. 10 m
- B. 10√3 m
- C. 20 m
- D. 20√3 m
Answer: A. 10 m — Let H be the height of the pole. In the right-angled triangle, sin 30° = H / 20. Since sin 30° = 1/2, H = 20 * (1/2) = 10 m.
MCQ 7. The angles of elevation of the top of a 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. The height of the tower is:
- A. 6 m
- B. 8 m
- C. 12 m
- D. 10 m
Answer: A. 6 m — Let H be the height of the tower. Let the angles be θ and (90-θ). tan θ = H/4 and tan (90-θ) = H/9. Since tan(90-θ) = cot θ = 1/tan θ, we get H/9 = 4/H, so H² = 36, and H = 6 m.
MCQ 8. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. The distance of the hill from the ship is:
- A. 10√3 m
- B. 20√3 m
- C. 30 m
- D. 10 m
Answer: A. 10√3 m — Let D be the distance of the hill from the ship. The angle of depression to the base is 30°, so tan 30° = 10/D. Thus, D = 10 / (1/√3) = 10√3 m.
MCQ 9. From a point on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagpole from the point is 45°. The length of the flagpole is:
- A. 10(√3 - 1) m
- B. 10√3 m
- C. 10(√3 + 1) m
- D. 5√3 m
Answer: A. 10(√3 - 1) m — Let H be the flagpole height and X the distance from building. tan 30° = 10/X => X = 10√3 m. tan 45° = (10+H)/X => 1 = (10+H)/(10√3) => 10√3 = 10+H => H = 10√3 - 10 = 10(√3 - 1) m.
MCQ 10. The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 20 meters towards the tower, the angle of elevation becomes 60°. The height of the tower is:
- A. 10 m
- B. 10√3 m
- C. 20 m
- D. 20√3 m
Answer: B. 10√3 m — Let H be the tower's height. Initial distance X, final distance (X-20). tan 30° = H/X and tan 60° = H/(X-20). Solving these gives H = 10√3 m.
MCQ 11. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, the distance between the two ships is:
- A. 75(√3 - 1) m
- B. 75√3 m
- C. 75/√3 m
- D. 75(√3 + 1) m
Answer: A. 75(√3 - 1) m — The distance of the closer ship (45°) is 75 m. The distance of the farther ship (30°) is 75√3 m. The distance between them is 75√3 - 75 = 75(√3 - 1) m.
MCQ 12. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. The height of the tower is:
- A. 42 m
- B. 38 m
- C. 40 m
- D. 32 m
Answer: A. 42 m — Due to similar triangles formed by the pole/tower and their shadows, the ratio of height to shadow is constant. So, 6/4 = Height_tower/28, which gives Height_tower = 42 m.
MCQ 13. Two persons are at a distance of 50 m from each other on opposite sides of a tower. The angles of elevation of the top of the tower are 30° and 60° respectively. The height of the tower is:
- A. 25√3/2 m
- B. 25√3 m
- C. 50√3 m
- D. 25/√3 m
Answer: A. 25√3/2 m — Let H be the tower's height. Let distances be x and (50-x). tan 60°=H/x and tan 30°=H/(50-x). Solving these equations for H gives H = 25√3/2 m.
MCQ 14. The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in the lake is β. The height of the cloud is:
- A. h(tan β + tan α) / (tan β - tan α)
- B. h(tan β - tan α) / (tan β + tan α)
- C. h(cot β + cot α) / (cot β - cot α)
- D. h(cot β - cot α) / (cot β + cot α)
Answer: A. h(tan β + tan α) / (tan β - tan α) — Using trigonometric ratios for the cloud (height H-h) and its reflection (height H+h) from the observer, we equate the horizontal distances to derive H = h(tan β + tan α) / (tan β - tan α).
MCQ 15. 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°. The height of the tower is:
- A. 7(1 + √3) m
- B. 7(1 - √3) m
- C. 7√3 m
- D. 14 m
Answer: A. 7(1 + √3) m — The distance to the tower (D) is found using tan 45° = 7/D, so D=7m. The height of the tower above the building is H-7. Using tan 60° = (H-7)/D, we get √3 = (H-7)/7, so H = 7 + 7√3 = 7(1+√3) m.
MCQ 16. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. The distance he walked towards the building is:
- A. 19√3 m
- B. 18√3 m
- C. 20√3 m
- D. 21√3 m
Answer: A. 19√3 m — The effective height of the building is 30 - 1.5 = 28.5 m. Let initial distance be X and final distance Y. X = 28.5√3 m and Y = 28.5/√3 = 9.5√3 m. The distance walked is X - Y = 19√3 m.
MCQ 17. 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. The distance from the foot of the tree to the point where the top touches the ground is 8 m. The original height of the tree is:
- A. 8√3 m
- B. 24 m
- C. 8(1 + √3) m
- D. 8(1 + 1/√3) m
Answer: A. 8√3 m — Let Y be the unbroken part and X be the broken part. tan 30° = Y/8 => Y = 8/√3 m. cos 30° = 8/X => X = 16/√3 m. Original height = X + Y = 24/√3 = 8√3 m.
MCQ 18. The angle of elevation of the top of a tower from two points on the ground, at distances 'a' and 'b' from the base of the tower and in the same straight line with it, are complementary. The height of the tower is:
- A. √(a/b)
- B. √(ab)
- C. √(a+b)
- D. √(a-b)
Answer: B. √(ab) — Let H be the height of the tower. Let the angles be θ and (90°-θ). Then tan θ = H/a and tan(90°-θ) = cot θ = H/b. Multiplying these gives tan θ * cot θ = (H/a) * (H/b), so 1 = H²/ab, which implies H = √ab.
MCQ 19. If the height of a vertical pole is equal to the length of its shadow on the ground, the angle of elevation of the sun is:
- A. 30°
- B. 45°
- C. 60°
- D. 90°
Answer: B. 45° — Let H be the height and S be the shadow. If H=S, then tan θ = H/S = 1. Therefore, the angle of elevation θ is 45°.
MCQ 20. The angles of elevation of the top of a tower from the base and top of a building of height 'h' are 60° and 30° respectively. The height of the tower is:
- A. h/2
- B. h√3
- C. 3h/2
- D. 2h
Answer: C. 3h/2 — Let H be the tower's height and X be the distance. tan 60° = H/X => X = H/√3. tan 30° = (H-h)/X. Substituting X, we get 1/√3 = (H-h)/(H/√3) => H = 3(H-h) => H = 3h/2.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.