NCERT Class 10 - Some Applications of Trigonometry (Exercise 9.1)

🎯 Key Concepts from Exercise 9.1

These are the fundamental principles you need to solve every problem in this exercise.

📐 Line of Sight

The line drawn from the eye of an observer to the point in the object viewed.
It forms the hypotenuse in most right triangle problems.

📏 Angle of Elevation

Angle formed by line of sight with horizontal when object is above eye level.
Used when looking UP at towers, kites, balloons, etc.

📉 Angle of Depression

Angle formed by line of sight with horizontal when object is below eye level.
Used when looking DOWN at ships, cars, foot of tower, etc.
Key: Angle of depression = Angle of elevation (alternate angles)

⚠️ Observer Height

Always subtract observer's height from total height when angle is measured from eye level.
Example: Boy (1.5m) looking at 30m building → Effective height = 28.5m

🌳 Broken Tree Problems

Original height = Standing part + Broken part (hypotenuse)
Broken part acts as the hypotenuse of the right triangle formed.

🔄 Relative Motion

For moving objects (cars, balloons), use uniform speed concept.
Distance = Speed × Time. Ratios help find unknown times.

📚 Essential Formulas

sin θ = Perpendicular / Hypotenuse | cos θ = Base / Hypotenuse | tan θ = Perpendicular / Base

Standard Values:
sin 30° = 1/2    cos 30° = √3/2    tan 30° = 1/√3
sin 45° = 1/√2    cos 45° = 1/√2    tan 45° = 1
sin 60° = √3/2    cos 60° = 1/2    tan 60° = √3

Key Identities:
tan θ = sin θ / cos θ | sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ

💡 Problem-Solving Strategy

1 Draw the diagram — Always sketch the situation first. Mark all known values.

2 Identify the right triangle — Find where the 90° angle is.

3 Choose the correct ratio — sin (opp/hyp), cos (adj/hyp), or tan (opp/adj).

4 Substitute values — Use standard angle values. Rationalize denominators.

5 Check units — Ensure all measurements are in the same unit (meters).

Trigonometry

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°.

✅ Step-by-Step Solution

1Let AB be the pole and AC be the rope.

2Given: AC = 20 m, ∠ACB = 30°

3In right ΔABC: sin 30° = AB/AC

4AB = AC × sin 30° = 20 × 1/2 = 10 m

5Therefore, the height of the pole is 10 m.

🎯 Final Answer: 10 m

Trigonometry

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.

✅ Step-by-Step Solution

1Let the tree break at point A, with standing part AB and broken part AC touching ground at C.

2Given: BC = 8 m, ∠ACB = 30°

3In right ΔABC: tan 30° = AB/BC ⇒ AB = 8 × tan 30° = 8/√3 m

4cos 30° = BC/AC ⇒ AC = 8/cos 30° = 16/√3 m

5Total height = AB + AC = 8/√3 + 16/√3 = 24/√3 = 8√3 m

6Therefore, the height of the tree is 8√3 m (≈ 13.86 m).

🎯 Final Answer: 8√3 m

Application

A contractor plans to install two slides. For children below 5 years: height 1.5 m, inclined at 30°. For older children: height 3 m, inclined at 60°. What should be the length of the slide in each case?

✅ Step-by-Step Solution

1For younger children: sin 30° = 1.5/L₁ ⇒ L₁ = 1.5/sin 30° = 1.5/(1/2) = 3 m

2For older children: sin 60° = 3/L₂ ⇒ L₂ = 3/sin 60° = 3/(√3/2) = 6/√3 = 2√3 m

3Therefore, lengths are 3 m and 2√3 m (≈ 3.46 m).

🎯 Final Answer: 3 m and 2√3 m

Trigonometry

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.

✅ Step-by-Step Solution

1Let AB be the tower and C be the point on ground.

2Given: BC = 30 m, ∠ACB = 30°

3In right ΔABC: tan 30° = AB/BC

4AB = 30 × tan 30° = 30 × (1/√3) = 30/√3 = 10√3 m

5Therefore, the height of the tower is 10√3 m (≈ 17.32 m).

🎯 Final Answer: 10√3 m

Trigonometry

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.

✅ Step-by-Step Solution

1Let A be the kite, C be the point on ground, and AC be the string.

2Given: AB = 60 m (height), ∠ACB = 60°

3In right ΔABC: sin 60° = AB/AC

4AC = AB/sin 60° = 60/(√3/2) = 120/√3 = 40√3 m

5Therefore, the length of the string is 40√3 m (≈ 69.28 m).

🎯 Final Answer: 40√3 m

Application

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.

✅ Step-by-Step Solution

1Height of building above boy's eyes = 30 - 1.5 = 28.5 m

2Initial distance (d₁): tan 30° = 28.5/d₁ ⇒ d₁ = 28.5√3 m

3Final distance (d₂): tan 60° = 28.5/d₂ ⇒ d₂ = 28.5/√3 m

4Distance walked = d₁ - d₂ = 28.5√3 - 28.5/√3

5= 28.5(√3 - 1/√3) = 28.5(2/√3) = 57/√3 = 19√3 m

6Therefore, he walked 19√3 m (≈ 32.91 m).

🎯 Final Answer: 19√3 m

Trigonometry

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.

✅ Step-by-Step Solution

1Let building height = 20 m, tower height = h, distance from point = d

2For bottom of tower: tan 45° = 20/d ⇒ d = 20 m

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

4√3 = (20+h)/20 ⇒ 20+h = 20√3

5h = 20√3 - 20 = 20(√3 - 1) m

6Therefore, the height of the tower is 20(√3-1) m (≈ 14.64 m).

🎯 Final Answer: 20(√3-1) m

Trigonometry

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.

✅ Step-by-Step Solution

1Let pedestal height = h, distance from point = d

2For top of pedestal: tan 45° = h/d ⇒ d = h

3For top of statue: tan 60° = (h+1.6)/d = (h+1.6)/h

4h√3 = h + 1.6 ⇒ h(√3 - 1) = 1.6

5h = 1.6/(√3-1) = 1.6(√3+1)/2 = 0.8(√3+1) m

6Therefore, the height of the pedestal is 0.8(√3+1) m (≈ 2.19 m).

🎯 Final Answer: 0.8(√3+1) m

Trigonometry

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.

✅ Step-by-Step Solution

1Let building height = h, distance between tower and building = d

2From foot of tower: tan 30° = h/d ⇒ d = h√3

3From foot of building: tan 60° = 50/d ⇒ d = 50/√3

4Therefore: h√3 = 50/√3

5h = 50/3 m

6Therefore, the height of the building is 50/3 m (≈ 16.67 m).

🎯 Final Answer: 50/3 m

Geometry

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.

✅ Step-by-Step Solution

1Let height of each pole = h, point P is at distance x from first pole and (80-x) from second.

2From first pole: tan 60° = h/x ⇒ h = x√3

3From second pole: tan 30° = h/(80-x) ⇒ h = (80-x)/√3

4Equating: x√3 = (80-x)/√3 ⇒ 3x = 80-x ⇒ 4x = 80 ⇒ x = 20 m

5Distance from second pole = 80-20 = 60 m

6Height h = 20√3 m (≈ 34.64 m)

7Therefore, height = 20√3 m and distances are 20 m and 60 m.

🎯 Final Answer: Height=20√3 m, Distances=20 m and 60 m

Geometry

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.

✅ Step-by-Step Solution

1Let tower height = h, width of canal = w

2From opposite bank: tan 60° = h/w ⇒ h = w√3

3From point 20 m away: tan 30° = h/(w+20) ⇒ h = (w+20)/√3

4Equating: w√3 = (w+20)/√3 ⇒ 3w = w+20 ⇒ 2w = 20 ⇒ w = 10 m

5Height h = 10√3 m (≈ 17.32 m)

6Therefore, height of tower = 10√3 m and width of canal = 10 m.

🎯 Final Answer: Height=10√3 m, Width=10 m

Trigonometry

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.

✅ Step-by-Step Solution

1Let tower height = H, distance between buildings = d

2Angle of depression to foot = 45°, so tan 45° = 7/d ⇒ d = 7 m

3Angle of elevation to top = 60°, so tan 60° = (H-7)/d = (H-7)/7

4√3 = (H-7)/7 ⇒ H-7 = 7√3

5H = 7 + 7√3 = 7(√3+1) m

6Therefore, the height of the tower is 7(√3+1) m (≈ 19.12 m).

🎯 Final Answer: 7(√3+1) m

Trigonometry

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.

✅ Step-by-Step Solution

1Let lighthouse = AB = 75 m. Ships at C (farther) and D (nearer).

2For ship D (45° depression): tan 45° = 75/BD ⇒ BD = 75 m

3For ship C (30° depression): tan 30° = 75/BC ⇒ BC = 75√3 m

4Distance between ships = BC - BD = 75√3 - 75 = 75(√3-1) m

5Therefore, the distance between the two ships is 75(√3-1) m (≈ 54.9 m).

🎯 Final Answer: 75(√3-1) m

Application

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.

✅ Step-by-Step Solution

1Height of balloon above girl's eyes = 88.2 - 1.2 = 87 m

2Initial horizontal distance: tan 60° = 87/d₁ ⇒ d₁ = 87/√3 = 29√3 m

3Later horizontal distance: tan 30° = 87/d₂ ⇒ d₂ = 87√3 m

4Distance travelled = d₂ - d₁ = 87√3 - 87/√3

5= 87(√3 - 1/√3) = 87(2/√3) = 174/√3 = 58√3 m

6Therefore, the distance travelled by the balloon is 58√3 m (≈ 100.46 m).

🎯 Final Answer: 58√3 m

Application

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.

✅ Step-by-Step Solution

1Let tower height = h, initial distance of car = d₁, distance after 6s = d₂

2tan 30° = h/d₁ ⇒ d₁ = h√3

3tan 60° = h/d₂ ⇒ d₂ = h/√3

4Distance covered in 6 seconds = d₁ - d₂ = h√3 - h/√3 = 2h/√3

5Speed of car = (2h/√3)/6 = h/(3√3)

6Time to reach foot from second point = d₂/speed = (h/√3) ÷ (h/(3√3)) = 3 seconds

7Therefore, the time taken is 3 seconds.

🎯 Final Answer: 3 seconds

📐 Interactive Right Triangle Visualizer

Adjust the angle and height to see how the triangle changes in real-time.

Angle (θ)

Height (m)

Scenario

Base: 173.2 m

Hypotenuse: 200.0 m

tan θ: 0.577

🌳 Broken Tree Visualizer

Visualize how a broken tree forms a right triangle.

Distance to top (m)

Angle with ground

Standing Part: 46.2 m

Total Height: 138.6 m

📐 Right Triangle Solver

Known Value

Angle (degrees)

Height (m)

Base: 17.32 m

Hypotenuse: 20.00 m

📏 Distance & Height Calculator

Angle of Elevation (°)

Distance from Object (m)

Observer Height (m) - optional

Object Height: 17.32 m

🔢 Trigonometric Values

Angle (degrees)

sin: 0.7071

cos: 0.7071

tan: 1.0000

🎯 Speed & Time (Q15 Type)

Initial Angle (°)

Final Angle (°)

Time Interval (seconds)

Time to reach: 3.0 sec

📐 Some Applications of Trigonometry

📐 Line of Sight

📏 Angle of Elevation

📉 Angle of Depression

⚠️ Observer Height

🌳 Broken Tree Problems

🔄 Relative Motion