April 10, 2026
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Chapter 9: Some Applications of Trigonometry
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
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Exercise 9.1 (Page 205)
13 marks
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Answer:
Let AB be the pole, AC the rope. ∠ACB = 30°, AC = 20 m.
sin 30° = AB/AC → 1/2 = AB/20 → AB = 10 m.
Height of pole = 10 m.
Let AB be the pole, AC the rope. ∠ACB = 30°, AC = 20 m.
sin 30° = AB/AC → 1/2 = AB/20 → AB = 10 m.
Height of pole = 10 m.
23 marks
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Answer:
Let original tree = AB, broken at C, top touches ground at D. ∠CDB=30°, DB=8 m.
In ΔCDB, tan30° = CB/DB → 1/√3 = CB/8 → CB = 8/√3 m.
cos30° = DB/CD → √3/2 = 8/CD → CD = 16/√3 m.
Height of tree = AB = AC + CB = CD + CB = 16/√3 + 8/√3 = 24/√3 = 8√3 ≈ 13.86 m.
Height = 8√3 m.
Let original tree = AB, broken at C, top touches ground at D. ∠CDB=30°, DB=8 m.
In ΔCDB, tan30° = CB/DB → 1/√3 = CB/8 → CB = 8/√3 m.
cos30° = DB/CD → √3/2 = 8/CD → CD = 16/√3 m.
Height of tree = AB = AC + CB = CD + CB = 16/√3 + 8/√3 = 24/√3 = 8√3 ≈ 13.86 m.
Height = 8√3 m.
33 marks
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Answer:
For younger children: height = 1.5 m, angle = 30°.
sin30° = height/length → 1/2 = 1.5/length → length = 3 m.
For elder children: height = 3 m, angle = 60°.
sin60° = 3/length → √3/2 = 3/length → length = 6/√3 = 2√3 ≈ 3.46 m.
Lengths: 3 m and 2√3 m.
For younger children: height = 1.5 m, angle = 30°.
sin30° = height/length → 1/2 = 1.5/length → length = 3 m.
For elder children: height = 3 m, angle = 60°.
sin60° = 3/length → √3/2 = 3/length → length = 6/√3 = 2√3 ≈ 3.46 m.
Lengths: 3 m and 2√3 m.
43 marks
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Answer:
Let tower height = h. tan30° = h/30 → 1/√3 = h/30 → h = 30/√3 = 10√3 ≈ 17.32 m.
Height = 10√3 m.
Let tower height = h. tan30° = h/30 → 1/√3 = h/30 → h = 30/√3 = 10√3 ≈ 17.32 m.
Height = 10√3 m.
53 marks
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Answer:
sin60° = height/length → √3/2 = 60/length → length = 120/√3 = 40√3 ≈ 69.28 m.
Length of string = 40√3 m.
sin60° = height/length → √3/2 = 60/length → length = 120/√3 = 40√3 ≈ 69.28 m.
Length of string = 40√3 m.
63 marks
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. Find the distance he walked towards the building.
Answer:
Height of building above eye level = 30 - 1.5 = 28.5 m.
Initial case: tan30° = 28.5/x → x = 28.5√3 m.
Final case: tan60° = 28.5/y → y = 28.5/√3 m.
Distance walked = x - y = 28.5√3 - 28.5/√3 = 28.5(√3 - 1/√3) = 28.5 × (2/√3) = 57/√3 = 19√3 ≈ 32.91 m.
Distance = 19√3 m.
Height of building above eye level = 30 - 1.5 = 28.5 m.
Initial case: tan30° = 28.5/x → x = 28.5√3 m.
Final case: tan60° = 28.5/y → y = 28.5/√3 m.
Distance walked = x - y = 28.5√3 - 28.5/√3 = 28.5(√3 - 1/√3) = 28.5 × (2/√3) = 57/√3 = 19√3 ≈ 32.91 m.
Distance = 19√3 m.
73 marks
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Answer:
Let building = AB = 20 m, tower = BC = h, point on ground D.
∠ADB = 45° → tan45° = AB/BD → 1 = 20/BD → BD = 20 m.
∠ADC = 60° → tan60° = (AB+BC)/BD → √3 = (20+h)/20 → 20√3 = 20+h → h = 20√3 - 20 = 20(√3 - 1) m.
Height of tower = 20(√3 - 1) m.
Let building = AB = 20 m, tower = BC = h, point on ground D.
∠ADB = 45° → tan45° = AB/BD → 1 = 20/BD → BD = 20 m.
∠ADC = 60° → tan60° = (AB+BC)/BD → √3 = (20+h)/20 → 20√3 = 20+h → h = 20√3 - 20 = 20(√3 - 1) m.
Height of tower = 20(√3 - 1) m.
83 marks
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 pedestal height = h, statue height = 1.6 m. Point distance = d.
tan45° = h/d → 1 = h/d → d = h.
tan60° = (h+1.6)/d → √3 = (h+1.6)/h → √3 h = h + 1.6 → h(√3 - 1) = 1.6 → h = 1.6/(√3 - 1).
Rationalize: h = 1.6(√3+1)/(3-1) = 1.6(√3+1)/2 = 0.8(√3+1) m.
Pedestal height = 0.8(√3+1) ≈ 2.19 m.
Let pedestal height = h, statue height = 1.6 m. Point distance = d.
tan45° = h/d → 1 = h/d → d = h.
tan60° = (h+1.6)/d → √3 = (h+1.6)/h → √3 h = h + 1.6 → h(√3 - 1) = 1.6 → h = 1.6/(√3 - 1).
Rationalize: h = 1.6(√3+1)/(3-1) = 1.6(√3+1)/2 = 0.8(√3+1) m.
Pedestal height = 0.8(√3+1) ≈ 2.19 m.
93 marks
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Answer:
Let building height = h. Let distance between them = d.
From tower foot: tan30° = h/d → 1/√3 = h/d → d = h√3.
From building foot: tan60° = 50/d → √3 = 50/d → d = 50/√3.
Equate: h√3 = 50/√3 → h = 50/3 ≈ 16.67 m.
Height of building = 50/3 m.
Let building height = h. Let distance between them = d.
From tower foot: tan30° = h/d → 1/√3 = h/d → d = h√3.
From building foot: tan60° = 50/d → √3 = 50/d → d = 50/√3.
Equate: h√3 = 50/√3 → h = 50/3 ≈ 16.67 m.
Height of building = 50/3 m.
103 marks
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:
Let height = h. Let point P, distance to one pole = x, to other = 80-x.
tan60° = h/x → √3 = h/x → h = x√3.
tan30° = h/(80-x) → 1/√3 = h/(80-x) → h = (80-x)/√3.
Equate: x√3 = (80-x)/√3 → 3x = 80 - x → 4x = 80 → x = 20 m.
Then h = 20√3 ≈ 34.64 m. Other distance = 60 m.
Height = 20√3 m, distances = 20 m and 60 m.
Let height = h. Let point P, distance to one pole = x, to other = 80-x.
tan60° = h/x → √3 = h/x → h = x√3.
tan30° = h/(80-x) → 1/√3 = h/(80-x) → h = (80-x)/√3.
Equate: x√3 = (80-x)/√3 → 3x = 80 - x → 4x = 80 → x = 20 m.
Then h = 20√3 ≈ 34.64 m. Other distance = 60 m.
Height = 20√3 m, distances = 20 m and 60 m.
113 marks
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Answer:
Let width of canal = x, height of tower = h.
First point: tan60° = h/x → √3 = h/x → h = x√3.
Second point is 20 m away from first (further from tower): distance from tower = x+20.
tan30° = h/(x+20) → 1/√3 = h/(x+20) → h = (x+20)/√3.
Equate: x√3 = (x+20)/√3 → 3x = x+20 → 2x = 20 → x = 10 m.
Then h = 10√3 m.
Width of canal = 10 m, height = 10√3 m.
Let width of canal = x, height of tower = h.
First point: tan60° = h/x → √3 = h/x → h = x√3.
Second point is 20 m away from first (further from tower): distance from tower = x+20.
tan30° = h/(x+20) → 1/√3 = h/(x+20) → h = (x+20)/√3.
Equate: x√3 = (x+20)/√3 → 3x = x+20 → 2x = 20 → x = 10 m.
Then h = 10√3 m.
Width of canal = 10 m, height = 10√3 m.
123 marks
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°. Determine the height of the tower.
Answer:
Let building AB = 7 m. Tower CD, with AB horizontal distance = x.
From top of building (point B), angle of depression to foot of tower = 45° → ∠CBD = 45° (alternate interior). In ΔBCD, ∠BCD=90°, ∠CBD=45° → BC = CD? Actually, in ΔBCD, tan45° = CD/BC → 1 = CD/BC → BC = CD.
Also, ∠ABD = 45° (angle of depression gives ∠EBD=45°, but careful: Let’s set correctly).
Standard approach: Let foot of tower = D, foot of building = B? Actually building AB, tower CD, with B and D at ground. From B, angle of depression to D is 45° → BD = AB = 7 m (since tan45° = AB/BD). So BD = 7 m.
From B, angle of elevation to C (top of tower) is 60° → tan60° = (CD - AB)/BD? No: height above building = (CD - 7)/7 → √3 = (CD-7)/7 → CD-7 = 7√3 → CD = 7(√3+1) m.
Height of tower = 7(√3+1) m ≈ 19.12 m.
Let building AB = 7 m. Tower CD, with AB horizontal distance = x.
From top of building (point B), angle of depression to foot of tower = 45° → ∠CBD = 45° (alternate interior). In ΔBCD, ∠BCD=90°, ∠CBD=45° → BC = CD? Actually, in ΔBCD, tan45° = CD/BC → 1 = CD/BC → BC = CD.
Also, ∠ABD = 45° (angle of depression gives ∠EBD=45°, but careful: Let’s set correctly).
Standard approach: Let foot of tower = D, foot of building = B? Actually building AB, tower CD, with B and D at ground. From B, angle of depression to D is 45° → BD = AB = 7 m (since tan45° = AB/BD). So BD = 7 m.
From B, angle of elevation to C (top of tower) is 60° → tan60° = (CD - AB)/BD? No: height above building = (CD - 7)/7 → √3 = (CD-7)/7 → CD-7 = 7√3 → CD = 7(√3+1) m.
Height of tower = 7(√3+1) m ≈ 19.12 m.
133 marks
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, find the distance between the two ships.
Answer:
Height = 75 m. For first ship (45° depression), distance from base = 75 / tan45° = 75 m.
For second ship (30° depression), distance = 75 / tan30° = 75√3 m.
Distance between ships = 75√3 - 75 = 75(√3 - 1) m.
Distance = 75(√3 - 1) m ≈ 54.9 m.
Height = 75 m. For first ship (45° depression), distance from base = 75 / tan45° = 75 m.
For second ship (30° depression), distance = 75 / tan30° = 75√3 m.
Distance between ships = 75√3 - 75 = 75(√3 - 1) m.
Distance = 75(√3 - 1) m ≈ 54.9 m.
143 marks
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Answer:
Girl's eye level height = 1.2 m. Balloon height above eyes = 88.2 - 1.2 = 87 m.
Initial angle 60° → horizontal distance d1 = 87 / tan60° = 87/√3 = 29√3 m.
Later angle 30° → horizontal distance d2 = 87 / tan30° = 87 * √3 = 87√3 m.
Distance travelled = d2 - d1 = 87√3 - 29√3 = 58√3 m.
Distance = 58√3 m ≈ 100.46 m.
Girl's eye level height = 1.2 m. Balloon height above eyes = 88.2 - 1.2 = 87 m.
Initial angle 60° → horizontal distance d1 = 87 / tan60° = 87/√3 = 29√3 m.
Later angle 30° → horizontal distance d2 = 87 / tan30° = 87 * √3 = 87√3 m.
Distance travelled = d2 - d1 = 87√3 - 29√3 = 58√3 m.
Distance = 58√3 m ≈ 100.46 m.
153 marks
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°, which is approaching the foot of the tower with a uniform speed. Six seconds later, 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 tower height = h. Let initial position of car be at distance x from foot, later at distance y.
tan30° = h/x → x = h√3.
tan60° = h/y → y = h/√3.
Distance covered in 6 seconds = x - y = h√3 - h/√3 = h(√3 - 1/√3) = h(2/√3).
Speed = (2h/√3)/6 = h/(3√3).
Remaining distance from later point to foot = y = h/√3.
Time = distance/speed = (h/√3) / (h/(3√3)) = (h/√3) × (3√3/h) = 3 seconds.
Time to reach foot = 3 seconds.
Let tower height = h. Let initial position of car be at distance x from foot, later at distance y.
tan30° = h/x → x = h√3.
tan60° = h/y → y = h/√3.
Distance covered in 6 seconds = x - y = h√3 - h/√3 = h(√3 - 1/√3) = h(2/√3).
Speed = (2h/√3)/6 = h/(3√3).
Remaining distance from later point to foot = y = h/√3.
Time = distance/speed = (h/√3) / (h/(3√3)) = (h/√3) × (3√3/h) = 3 seconds.
Time to reach foot = 3 seconds.
163 marks
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. Prove that the height of the tower is 6 m.
Answer:
Let height = h. Let angle at 4 m = θ, then at 9 m = 90° - θ.
tanθ = h/4, tan(90° - θ) = cotθ = h/9.
Multiply: (h/4) × (h/9) = tanθ × cotθ = 1 → h² = 36 → h = 6 m (positive).
Hence proved.
Let height = h. Let angle at 4 m = θ, then at 9 m = 90° - θ.
tanθ = h/4, tan(90° - θ) = cotθ = h/9.
Multiply: (h/4) × (h/9) = tanθ × cotθ = 1 → h² = 36 → h = 6 m (positive).
Hence proved.
Key Concepts Summary
FormulaTrigonometric Ratios
What are the three primary trigonometric ratios used in heights and distances?
Answer: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
Angle of Elevation/DepressionDefinition
Define angle of elevation and angle of depression.
Answer: Angle of elevation is the angle from the horizontal upward to an object. Angle of depression is the angle from the horizontal downward to an object.
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