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NCERT Solutions Class 10 Maths Chapter 9 Exercise 9.1 Some Applications of Trigonometry

NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry Exercise 9.1 are designed to help students apply trigonometric ratios to solve real-life problems. This exercise introduces the practical applications of trigonometry in calculating heights and distances, which is a crucial part of geometry.

Exercise 9.1 focuses on:

NCERT Solutions Class 10 Maths Chapter 9 Exercise 9.1 Some Applications of Trigonometry

Some Applications of Trigonometry

Medium

Q.

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° (see the following figure).

Some Applications of Trigonometry

Medium

Q.

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.

Some Applications of Trigonometry

Medium

Q.

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.

Some Applications of Trigonometry

Medium

Q.

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° (see the following figure). Find the height of the tower and the width of the canal.

Some Applications of Trigonometry

Medium

Q.

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.

Some Applications of Trigonometry

Medium

Q.

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.

Some Applications of Trigonometry

Medium

Q.

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° (see the following figure). Find the distance travelled by the balloon during the interval.

NCERT Solutions Class 10 Maths Chapter 9 Exercise 9.1 Some Applications of Trigonometry

Solving problems involving heights and distances using trigonometric ratios (sin, cos, and tan).
Finding the height of buildings, distance between objects, and angles of elevation and depression in real-world contexts.
Applying trigonometric identities to simplify and solve problems.

The solutions provide a clear, step-by-step approach that helps students understand how to use trigonometric ratios in practical problems, making it easier for them to tackle such questions in their Class 10 exams.

Q1. 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° (see Fig. 9.11).

Difficulty Level: Easy
Known:
(i) Length of rope = 20 m
(ii) Angle of rope with ground = 30° = ∠ACB

Unknown:
Height of pole

Reasoning:
AB = Height of the Pole
BC = Distance between the point on the ground and the pole.
AC = Length of the Rope (Hypotenuse)

We need to find the height of the pole AB, from the angle C and the length of the rope AC. Therefore, Trigonometric ratio involving all the three measures is sin C.
In △ABC,

$\sin C = \frac{AB}{AC}$

$sin C = A C A B$

$\sin 30^\circ = \frac{AB}{20}$

$sin 3 0^{\circ} = 20 A B$

$\frac{1}{2} = \frac{AB}{20}$

$21 = 20 A B$

$AB = 20 \times \frac{1}{2} = 10 \, \text{m}$

$A B = 20 \times 21 = 10 m$

Answer:
Height of pole

$AB = 10 \, \text{m}$

$A B = 10 m$

Q2. 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.

Difficulty Level: Medium
Unknown:
Height of the tree

Known:
(i) Broken part of the tree bends and touches the ground making an angle of 30° with the ground.
(ii) Distance between foot of the tree to the top of the tree is 8m

Reasoning:
(i) Height of the tree = AB + AC
(ii) Trigonometric ratio which involves AB, BC, and ∠C is tan θ, where AB can be measured.
(iii) Trigonometric ratio which involves AB, AC, and ∠C is sin θ, where AC can be measured.
(iv) Distance between the foot of the tree to the point where the top touches the ground = BC = 8 m

Solution:
In △ABC,

$\tan 30^\circ = \frac{AB}{BC}$

$tan 3 0^{\circ} = BC A B$

$\tan 30^\circ = \frac{AB}{8}$

$tan 3 0^{\circ} = 8 A B$

$\frac{1}{\sqrt{3}} = \frac{AB}{8}$

$3 1 = 8 A B$

$AB = \frac{8}{\sqrt{3}} \approx 4.62 \, \text{m}$

$A B = 3 8 \approx 4.62 m$

For the broken part of the tree, we can use sin 30°:

$\sin 30^\circ = \frac{AC}{8}$

$sin 3 0^{\circ} = 8 A C$

$\frac{1}{2} = \frac{AC}{8}$

$21 = 8 A C$

$AC = 4 \, \text{m}$

$A C = 4 m$

Total height of the tree =

$AB + AC = 4.62 + 4 = 8.62 \, \text{m}$

$A B + A C = 4.62 + 4 = 8.62 m$

Answer:
Height of the tree = 8.62 m

Q3. 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 3m and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Difficulty Level: Medium
Unknown:
Length of the slide for children below the age of 5 years and elder children.

Known:
(i) For the children below the age of 5 years:
Height of the slide = 1.5 m
Slide’s angle with the ground = 30°

(ii) For elder children:
Height of the slide = 3 m
Slide’s angle with the ground = 60°

Reasoning:
Let us consider the following conventions for the slide installed for children below 5 years:

The height of the slide as AC.
Distance between the foot of the slide to the point where it touches the ground as AB.
Length of the slide as BC

Let us consider the following conventions for the slide installed for elder children:

The height of the slide PR.
Distance between the foot of the slide to the point where it touches the ground as PQ.
Length of the slide as QR

(i) Trigonometric ratio involving AC, BC, and ∠B is sin θ
(ii) Trigonometric ratio involving PR, QR, and ∠Q is sin θ

Solution:
(i) In △ABC,

$\sin 30^\circ = \frac{AC}{BC}$

$sin 3 0^{\circ} = BC A C$

$\sin 30^\circ = \frac{1.5}{BC}$

$sin 3 0^{\circ} = BC 1.5$

$\frac{1}{2} = \frac{1.5}{BC}$

$21 = BC 1.5$

$BC = 3 \, \text{m}$

$BC = 3 m$

(ii) In △PRQ,

$\sin 60^\circ = \frac{PR}{QR}$

$sin 6 0^{\circ} = QR PR$

$\sin 60^\circ = \frac{3}{QR}$

$sin 6 0^{\circ} = QR 3$

$\frac{\sqrt{3}}{2} = \frac{3}{QR}$

$2 3 = QR 3$

$QR = \frac{3}{\frac{\sqrt{3}}{2}} = 2\sqrt{3} \approx 3.464 \, \text{m}$

$QR = 2 3 3 = 23 \approx 3.464 m$

Answer:
Length of slide for children below 5 years = 3 m
Length of slide for elder children = 3.464 m

Q4. 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.

Difficulty Level: Medium
Known:
(i) Angle of elevation of the top of the tower from a point on ground is 30°
(ii) Distance between the foot of the tower to the point on the ground is 30 m.

Unknown:
Height of the tower

Reasoning:
Let us consider the height of the tower as AB, distance between the foot of the tower to the point on ground as BC.
In △ABC,
Trigonometric ratio involving AB, BC and ∠C is tan θ.

Solution:
In △ABC,

$\tan 30^\circ = \frac{AB}{BC}$

$tan 3 0^{\circ} = BC A B$

$\tan 30^\circ = \frac{AB}{30}$

$tan 3 0^{\circ} = 30 A B$

$\frac{1}{\sqrt{3}} = \frac{AB}{30}$

$3 1 = 30 A B$

$AB = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \, \text{m}$

$A B = 3 30 = 103 \approx 17.32 m$

Answer:
Height of tower

$AB = 10\sqrt{3} \, \text{m}$

$A B = 103 m$

Q5. 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.

Difficulty Level: Medium
Known:
(i) Height of the flying kite = 60 m
(ii) Angle made by the string to the ground = 60°

Unknown:
Length of the string

Reasoning:
Let the height of the flying kite as AB, length of the string as AC and the inclination of the string with the ground as ∠C.
Trigonometric ratio involving AB, AC and ∠C is sin θ.

Solution:
In △ABC,

$\sin 60^\circ = \frac{AB}{AC}$

$sin 6 0^{\circ} = A C A B$

$\sin 60^\circ = \frac{60}{AC}$

$sin 6 0^{\circ} = A C 60$

$\frac{\sqrt{3}}{2} = \frac{60}{AC}$

$2 3 = A C 60$

$AC = \frac{60}{\frac{\sqrt{3}}{2}} = 40\sqrt{3} \approx 69.28 \, \text{m}$

$A C = 2 3 60 = 403 \approx 69.28 m$

Answer:
Length of the string

$AC = 40\sqrt{3} \, \text{m}$

$A C = 403 m$

FAQs: Class 10 Maths Chapter 9 – Some Applications of Trigonometry Exercise 9.1

Q1. What is the focus of Exercise 9.1?
Answer:
Exercise 9.1 focuses on applying trigonometric ratios (sin, cos, tan) to solve problems related to heights and distances, and understanding the angles of elevation and depression in real-world scenarios.

Q2. What is the angle of elevation?
Answer:
The angle of elevation is the angle formed by the line of sight of an observer when they look up at an object. It is measured from the horizontal line to the object.

Q3. What is the angle of depression?
Answer:
The angle of depression is the angle formed by the line of sight of an observer when they look down at an object. It is measured from the horizontal line to the object.

Q4. How do I apply trigonometric ratios to find the height of an object?
Answer:
To find the height of an object:

Identify the distance from the object and the angle of elevation or depression.
Use the appropriate trigonometric ratio (tan, sin, cos) based on the given information.
Apply the formula and solve for the unknown.

For example, if you know the distance and the angle of elevation, use:

$\tan \theta = \frac{\text{height}}{\text{distance}}$

$tan θ = distance height$

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide step-by-step explanations and practical examples of using trigonometric ratios to solve real-life problems. This improves students' problem-solving skills and helps them prepare efficiently for board exams.