NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 2026-27

NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry help students understand how trigonometry is used to solve practical problems related to heights and distances. In this chapter, students learn how to find the height of a building, tower, tree, or pole and how to calculate the distance between two points without measuring them directly. These calculations are done with the help of trigonometric ratios, angles of elevation, and angles of depression.

The chapter connects classroom mathematics with real-life situations. For example, students may need to find the height of a tower by observing its top from a certain distance or calculate the width of a river without crossing it. By understanding the given information, drawing a correct figure, and choosing the right trigonometric ratio, such questions can be solved easily.

NCERT Solutions for Class 10 Maths Chapter 9 provide clear and step-by-step answers to all the exercises given in the textbook. Each solution explains the method in simple words so that students can understand why a particular formula or ratio is used. The solutions also help students avoid common mistakes while drawing diagrams, identifying angles, and selecting values.

Infinity Learn offers well-explained NCERT Solutions that make learning easier and more organised. These solutions are useful for completing homework, revising important concepts, practising exam questions, and preparing for board examinations. Students can use them to check their answers and improve their problem-solving skills.

With regular practice, students can become confident in solving application-based trigonometry questions. Overall, NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry are a helpful study resource for understanding the chapter clearly and performing better in exams. They also support quick revision before tests and help learners build a strong base for studying advanced trigonometry successfully in higher classes.

Download Class 10 Maths Chapter 9 Some Applications of Trigonometry NCERT Solutions PDF

Students can download the Class 10 Maths Chapter 9 Some Applications of Trigonometry NCERT Solutions PDF to study the chapter easily at any time. The PDF contains simple and step-by-step solutions to the questions given in the NCERT textbook. It is a useful study resource for understanding important concepts, completing homework, revising the chapter, and preparing for school and board examinations.

Access NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Question 1: A circus performer climbs a tightly stretched rope of length 20 m. The rope is tied between the top of a vertical pole and a point on the ground. It forms an angle of 30° with the ground. Find the height of the pole.

Solution:

Let:

• AB represent the vertical pole.
• AC represent the rope.
• BC represent the horizontal ground.

Here, AC = 20 m and ∠ACB = 30°.

In the right-angled triangle ABC, AB is opposite to the 30° angle and AC is the hypotenuse.

sin 30° = AB/AC

1/2 = AB/20

AB = 20 × 1/2

AB = 10 m

Therefore, the pole is 10 m high.

Question 2: A tree breaks during a storm. Its broken upper part bends and touches the ground at a point 8 m away from the foot of the tree. The broken part makes an angle of 30° with the ground. Find the original height of the tree.

Solution:

Let:

• BC be the unbroken vertical part of the tree.
• AB be the broken upper part.
• AC = 8 m be the distance from the foot of the tree to the point where its top touches the ground.

The broken part AB makes an angle of 30° with the ground.

Step 1: Find the height of the remaining stump

In right-angled triangle ABC:

tan 30° = BC/AC

1/√3 = BC/8

BC = 8/√3 m

Step 2: Find the length of the broken part

cos 30° = AC/AB

√3/2 = 8/AB

AB = (8 × 2)/√3

AB = 16/√3 m

Step 3: Find the original height

Original height = BC + AB

Original height = 8/√3 + 16/√3

Original height = 24/√3

Original height = (24√3)/3

Original height = 8√3 m

Therefore, the original height of the tree was 8√3 m.

Question 3: A contractor wants to install two slides in a park. The slide for children below five years of age has a vertical height of 1.5 m and is inclined at 30° to the ground. The slide for older children has a vertical height of 3 m and is inclined at 60° to the ground. Find the length of each slide.

Solution:

Slide for younger children

Let the length of the first slide be L₁.

The vertical height is 1.5 m and the angle with the ground is 30°.

sin 30° = Vertical height/Length of slide

1/2 = 1.5/L₁

L₁ = 1.5 × 2

L₁ = 3 m

Slide for older children

Let the length of the second slide be L₂.

The vertical height is 3 m and the angle with the ground is 60°.

sin 60° = Vertical height/Length of slide

√3/2 = 3/L₂

L₂ = (3 × 2)/√3

L₂ = 6/√3

L₂ = 2√3 m

Therefore, the lengths of the two slides are 3 m and 2√3 m.

Question 4: A point on the ground is 30 m away from the foot of a tower. The angle of elevation of the top of the tower from this point is 30°. Find the height of the tower.

Solution:

Let:

• AB be the tower.
• BC = 30 m be the horizontal distance from the observation point to the tower.
• ∠ACB = 30°.

In right-angled triangle ABC:

tan 30° = AB/BC

1/√3 = AB/30

AB = 30/√3

AB = (30√3)/3

AB = 10√3 m

Therefore, the height of the tower is 10√3 m.

Question 5: A kite is flying 60 m above the ground. Its string is tied to a point on the ground and makes an angle of 60° with the horizontal ground. Assuming that the string is completely stretched, find its length.

Solution:

Let:

• KL = 60 m be the vertical height of the kite.
• KP be the length of the string.
• ∠KPL = 60°.

In right-angled triangle KLP, KP is the hypotenuse.

sin 60° = KL/KP

√3/2 = 60/KP

KP = (60 × 2)/√3

KP = 120/√3

KP = 40√3 m

Therefore, the string is 40√3 m long.

Question 6: A boy whose eye level is 1.5 m above the ground stands at a certain distance from a 30 m high building. As he moves towards the building, the angle of elevation of its top changes from 30° to 60°. Find the distance walked by the boy.

Solution:

The height used in the right-angled triangles is the difference between the height of the building and the boy's eye level.

Effective vertical height = 30 - 1.5

Effective vertical height = 28.5 m

Effective vertical height = 57/2 m

Step 1: Find the initial distance

Let the initial horizontal distance from the building be x.

The initial angle of elevation is 30°.

tan 30° = (57/2)/x

1/√3 = (57/2)/x

x = (57√3)/2 m

Step 2: Find the distance after walking forward

Let the new horizontal distance from the building be y.

The new angle of elevation is 60°.

tan 60° = (57/2)/y

√3 = (57/2)/y

y = 57/(2√3)

y = 19√3/2 m

Step 3: Calculate the distance walked

Distance walked = x - y

Distance walked = (57√3)/2 - (19√3)/2

Distance walked = (38√3)/2

Distance walked = 19√3 m

Therefore, the boy walked 19√3 m towards the building.

Question 7: A transmission tower is installed on the roof of a building that is 20 m high. From a point on the ground, the angles of elevation of the bottom and top of the tower are 45° and 60°, respectively. Find the height of the transmission tower.

Solution:

Let:

• BC = 20 m be the height of the building.
• AB be the height of the tower.
• CD be the horizontal distance from the observation point to the building.

Step 1: Find the horizontal distance

The angle of elevation of the bottom of the tower, which is the top of the building, is 45°.

tan 45° = BC/CD

1 = 20/CD

CD = 20 m

Step 2: Find the total height of the building and tower

The angle of elevation of the top of the tower is 60°.

tan 60° = (AB + BC)/CD

√3 = (AB + 20)/20

20√3 = AB + 20

AB = 20√3 - 20

AB = 20(√3 - 1) m

Therefore, the transmission tower is 20(√3 - 1) m high.

Question 8: A statue measuring 1.6 m in height is placed on a pedestal. From a point on the ground, the angle of elevation of the top of the pedestal is 45°, while the angle of elevation of the top of the statue is 60°. Find the height of the pedestal.

Solution:

Let:

• AB = 1.6 m be the height of the statue.
• BC = h m be the height of the pedestal.
• CD = x m be the horizontal distance from the observation point.

Step 1: Use the 45° angle

tan 45° = BC/CD

1 = h/x

x = h

Step 2: Use the 60° angle

The total height up to the top of the statue is h + 1.6.

tan 60° = (h + 1.6)/x

√3 = (h + 1.6)/h

h√3 = h + 1.6

h√3 - h = 1.6

h(√3 - 1) = 1.6

h = 1.6/(√3 - 1)

Step 3: Rationalise the denominator

h = [1.6(√3 + 1)]/[(√3 - 1)(√3 + 1)]

h = [1.6(√3 + 1)]/(3 - 1)

h = [1.6(√3 + 1)]/2

h = 0.8(√3 + 1) m

Therefore, the pedestal is 0.8(√3 + 1) m high.

Question 9: The angle of elevation of the top of a building from the foot of a tower is 30°. 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.

Solution:

Let:

• AB = h m be the height of the building.
• CD = 50 m be the height of the tower.
• BD = x m be the distance between them.

Step 1: Find the distance between the building and tower

From the foot of the building, the angle of elevation of the tower's top is 60°.

tan 60° = CD/BD

√3 = 50/x

x = 50/√3 m

Step 2: Find the height of the building

From the foot of the tower, the angle of elevation of the building's top is 30°.

tan 30° = AB/BD

1/√3 = h/(50/√3)

h = (50/√3) × (1/√3)

h = 50/3 m

h = 16 2/3 m

Therefore, the height of the building is 50/3 m, or 16 2/3 m.

Question 10: Two poles of the same height stand on opposite sides of an 80 m wide road. From a point between the poles, the angles of elevation of their tops are 60° and 30°. Find the height of each pole and the distances of the point from the two poles.

Solution:

Let:

• AB and CD be the two equal poles.
• The common height of each pole be h m.
• O be the observation point.
• BO = x m.
• DO = 80 - x m.

Suppose the angle of elevation of the first pole is 60° and that of the second pole is 30°.

Step 1: Form an equation using the 60° angle

tan 60° = h/x

√3 = h/x

h = x√3

Step 2: Form an equation using the 30° angle

tan 30° = h/(80 - x)

1/√3 = h/(80 - x)

h = (80 - x)/√3

Step 3: Use the fact that both poles have equal heights

x√3 = (80 - x)/√3

3x = 80 - x

4x = 80

x = 20 m

Therefore, the distance from the pole seen at 60° is 20 m.

Distance from the other pole = 80 - 20

Distance from the other pole = 60 m

Step 4: Find the height

h = x√3

h = 20√3 m

Therefore, each pole is 20√3 m high. The observation point is 20 m from one pole and 60 m from the other.

Question 11: A television tower stands vertically on one bank of a canal. From a point on the opposite bank, directly across from the tower, the angle of elevation of its top is 60°. From another point located 20 m farther away along the same straight line, the angle of elevation is 30°. Find the height of the tower and the width of the canal.

Solution:

Let:

• AB = h m be the height of the tower.
• BC = x m be the width of the canal.
• CD = 20 m.

Thus, the distance from the second observation point to the tower is x + 20 m.

Step 1: Use the angle of 60°

tan 60° = AB/BC

√3 = h/x

h = x√3

Step 2: Use the angle of 30°

tan 30° = AB/(BC + CD)

1/√3 = h/(x + 20)

h√3 = x + 20

Step 3: Substitute h = x√3

(x√3) × √3 = x + 20

3x = x + 20

2x = 20

x = 10 m

Therefore, the width of the canal is 10 m.

Step 4: Find the tower's height

h = x√3

h = 10√3 m

Therefore, the tower is 10√3 m high and the canal is 10 m wide.

Question 12: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60°, while the angle of depression of the foot of the tower is 45°. Find the height of the cable tower.

Solution:

Let:

• AB = 7 m be the height of the building.
• CD be the height of the cable tower.
• BD be the horizontal distance between the building and the tower.
• E be a point on the tower at the same horizontal level as A.

Since A and E are at the same level:

ED = AB = 7 m

Also, AE = BD.

Step 1: Find the horizontal distance

The angle of depression of D from A is 45°. Therefore, the angle of elevation of A from D is also 45°.

tan 45° = AB/BD

1 = 7/BD

BD = 7 m

Therefore, AE = 7 m.

Step 2: Find the part of the tower above the building

tan 60° = CE/AE

√3 = CE/7

CE = 7√3 m

Step 3: Find the total height of the tower

CD = CE + ED

CD = 7√3 + 7

CD = 7(√3 + 1) m

Therefore, the cable tower is 7(√3 + 1) m high.

Question 13: A lighthouse is 75 m high. From its top, the angles of depression of two ships on the same side are 45° and 30°. One ship is directly behind the other. Find the distance between the ships.

Solution:

Let:

• AB = 75 m be the lighthouse.
• C be the position of the nearer ship.
• D be the position of the farther ship.

The angles of elevation from the ships are equal to the corresponding angles of depression from the top of the lighthouse.

Step 1: Find the distance of the nearer ship

The angle for the nearer ship is 45°.

tan 45° = AB/BC

1 = 75/BC

BC = 75 m

Step 2: Find the distance of the farther ship

The angle for the farther ship is 30°.

tan 30° = AB/BD

1/√3 = 75/BD

BD = 75√3 m

Step 3: Find the distance between the ships

CD = BD - BC

CD = 75√3 - 75

CD = 75(√3 - 1) m

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

Question 14: A girl is 1.2 m tall and observes a balloon moving horizontally at a height of 88.2 m above the ground. At first, the angle of elevation of the balloon from her eyes is 60°. After some time, the angle becomes 30°. Find the horizontal distance travelled by the balloon.

Solution:

The vertical height of the balloon above the girl's eye level is:

Vertical height = 88.2 - 1.2

Vertical height = 87 m

Step 1: Find the initial horizontal distance

Let the initial horizontal distance be x.

tan 60° = 87/x

√3 = 87/x

x = 87/√3

x = 29√3 m

Step 2: Find the later horizontal distance

Let the later horizontal distance be y.

tan 30° = 87/y

1/√3 = 87/y

y = 87√3 m

Step 3: Calculate the distance travelled

Distance travelled = y - x

Distance travelled = 87√3 - 29√3

Distance travelled = 58√3 m

Therefore, the balloon travels 58√3 m.

Question 15: A straight road leads to the foot of a tower. A man standing at the top of the tower sees a car approaching the tower at a constant speed. Initially, the angle of depression of the car is 30°. Six seconds later, the angle of depression becomes 60°. Find the additional time required by the car to reach the foot of the tower.

Solution:

Let:

• AB = h m be the height of the tower.
• C be the initial position of the car.
• D be its position after six seconds.
• B be the foot of the tower.

The angles of elevation from the car are equal to the corresponding angles of depression from the top of the tower.

Step 1: Find the initial distance of the car

At C, the angle of elevation is 30°.

tan 30° = AB/BC

1/√3 = h/BC

BC = h√3

Step 2: Find the distance after six seconds

At D, the angle of elevation is 60°.

tan 60° = AB/BD

√3 = h/BD

BD = h/√3

Step 3: Find the distance covered in six seconds

CD = BC - BD

CD = h√3 - h/√3

CD = (3h - h)/√3

CD = 2h/√3

The car covers 2h/√3 m in 6 seconds.

Step 4: Find the time required to cover the remaining distance

The remaining distance is:

BD = h/√3

The remaining distance is exactly half of the distance covered during the previous six seconds:

(h/√3) ÷ (2h/√3) = 1/2

Because the car moves at a constant speed:

Required time = 6 × 1/2

Required time = 3 seconds

Therefore, the car will take 3 more seconds to reach the foot of the tower.