NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry 

Some Applications of Trigonometry focuses on real-life applications of trigonometry, particularly in measuring heights and distances without the actual need to measure them physically. This chapter introduces students to concepts like angles of elevation and depression and how they can be used to solve problems in day-to-day life, such as determining the height of a building or the distance of a ship from a lighthouse.

Some applications of trigonometry Class 10 NCERT Solutions provide step-by-step explanations for all the exercise questions, helping students understand the application of trigonometric concepts in solving practical problems. NCERT Solutions Class 10 Maths are prepared by ALLEN’s experienced educators to ensure that they are easy to understand and follow.

To score good marks in exams, students should regularly practice the NCERT Solutions and ensure they have a strong grasp of the concepts. Utilizing some applications of trigonometry class 10 notes alongside these solutions will further enhance their understanding and problem-solving skills, allowing them to tackle questions with confidence and accuracy.    

1.0Download NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry : Free PDF

2.0NCERT Solutions Class 10 Maths Chapter 9 Some applications of Trigonometry: All Exercises

In order to assure good scores in examinations, students need to practice all the exercises in some applications of trigonometry class 10 solutions on a regular basis and have a proper understanding of the concepts. These applications of Class 10 Trigonometry Notes, which come along with the solutions, help in enhancing their understanding and ability to solve problems and thus the students can answer questions sharply and accurately.

3.0What Will Students Learn in Chapter 9: Some Applications of Trigonometry?

• Introduction to Trigonometric Ratios.
• Application of Trigonometric Ratios in determining heights and distances.
• Concept of angle of elevation and angle of depression.
• Practical problems based on real-life situations involving trigonometry.

4.0Important Topics of Class 10 Math Chapter 9 Application of Trigonometry

Chapter 9, Application of Trigonometry, focuses on using trigonometric ratios to solve real-life problems involving heights, distances, and angles. Key concepts include the angle of elevation and angle of depression, along with applying sine, cosine, and tangent ratios in right-angled triangles. This chapter helps students calculate difficult-to-measure distances and heights in practical scenarios, such as surveying, navigation, and architecture, providing essential skills for solving real-world trigonometric problems.

1. Heights and Distances
2. Angle of Elevation
3. Angle of Depression

5.0General Outline for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Understanding Key Concepts:

Real-World Applications: The key idea is to apply trigonometric ratios (sine, cosine, tangent) to solve problems involving heights, distances, and angles in practical situations.

Types of Problems:

• Angle of Elevation: The angle formed between the line of sight of an observer and the horizontal ground when looking up at an object.
• Angle of Depression: The angle formed between the line of sight of an observer and the horizontal ground when looking down at an object.

Trigonometric Ratios:

• Sine $(\sin \theta )=\frac{\text{Opposite}}{\text{Hypotenuse}}$
• Cosine $(\cos \theta )=\frac{\text{Adjacent}}{\text{Hypotenuse}}$
• Tangent $(\tan \theta )=\frac{\text{Opposite}}{\text{Adjacent}}$

Common Problem Types:

Finding the Height of an Object:

• Given the angle of elevation and the distance from the object, use $\tan \theta$ to find the height.
• Example: A person standing at a distance from a building sees the top of the building at an angle of elevation of 30°. Find the height of the building.

Finding the Distance Between Two Points:

• Given the height and angle of elevation/depression, use the appropriate trigonometric ratio to calculate the distance.
• Example: A tower is 100 m tall. The angle of depression to a point on the ground is 45°. Find the distance between the point and the tower.

Finding the Angle of Elevation/Depression:

• Given the sides of the triangle (height and distance), use inverse trigonometric functions $({\sin }^{-1},{\cos }^{-1},{\tan }^{-1})$ to find the angle.
• Example: A person is standing 50 meters away from a building and sees the top at an angle of elevation of 60°. Find the height of the building.

Key Formulas:

• Height of an Object (using tangent): $\tan \theta =\frac{\text{Height}}{\text{Distance}}$
• Distance to an Object (using sine or cosine): $\sin \theta =\frac{\text{Height}}{\text{Hypotenuse}}\text{or}\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}$
• Using Inverse Trigonometric Functions: $\theta ={\tan }^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right),\theta ={\sin }^{-1}\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right)$

6.0Sample NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

1. 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.).

Solution:

AC = 20 m is the length of the rope.

Let AB = h meters be the height of the pole.

∠ACB = 30° (Given)

Now,

AB/AC = sin 30° = 1/2

=> h/20 = 1/2

=> h = 10 m

∴ the height of the pole is 10 m.

2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. 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.

Solution:

Let a tree be broken at point A, and its top be touching the ground at point B.

Now, in right triangle AOB, we have:

AO/OB = tan 30°

=> AO/8 = 1/√3

=> AO = 8/√3 meters

Also, AB/OB = sec 30°

=> AB/8 = 2/√3

=> AB = (2 * 8)/√3 = 16/√3 meters

Now, the height of the tree OP = OA + AP

= OA + AB

= 8/√3 + 16/√3 = 24/√3 * (√3/√3) meters = 8√3 meters

Therefore, the height of the tree is 8√3 meters.

3. A contractor plans to install two slides for children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 meters 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 meters and is inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

In the figure, $l_1$ is the length of the slide made for children below the age of 5 years, and $l_2$ is the length of the slide made for older children.

(for children below 5 years)

In the given figure, AB = 1.5 m, AC = l1 m, and ∠ACB = 30°, PQ = 3 m, OP = l2 m, and ∠POQ = 60°.

We have:

AB / AC = sin 30°

PQ / OP = sin 60°

Substituting the given values:

1.5 / l1 = 1/2

3 / l2 = √3 / 2

Solving for l1:

l1 = 2 × 1.5 m

l1 = 3 m

Solving for l2:

l2 = (3 × 2) / √3 m

l2 = 6 / √3 m

l2 = (6√3) / 3 m

l2 = 2√3 m

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

4. 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.

Solution:

Let AB be the height of the tower, denoted as h. Let C be the point on the ground which is 30 m away from the foot of the tower (B). We are given that ∠ACB = 30°.

Therefore, AC = 30 m.

Now, AB/AC = tan 30°

=> h/30 = 1/√3

=> h = 30/√3 = (30/√3) * (√3/√3) = 10√3

Thus, the required height of the tower is 10√3 m.

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°. Find the length of the string, assuming that there is no slack in the string.

Solution:

P is the position of the kite. Its height from the point Q (on the ground) = PQ = 60 m.

Let OP = l be the length of the string.

∠POQ = 60° (Given)

Now, PQ/OP = sin 60°

=> 60/l = sin 60° = √3/2

=> 60/l = √3/2

=> l = 40√3 m

Therefore, the length of the string is 40√3 m.

7.0Benefits of Some Applications of Trigonometry

The Some Applications of Trigonometry chapter in Class 10 Maths offers several key benefits:

1. Real-World Applications: It helps in solving practical problems related to heights and distances, which are commonly encountered in fields like architecture, engineering, and navigation.
2. Skill Development: Students learn to apply trigonometric ratios to calculate unknown distances and angles, improving problem-solving skills.
3. Understanding Geometry: The chapter reinforces the connection between trigonometry and geometry, particularly in the context of triangles.
4. Practical Use in Science: It builds a foundation for applying trigonometric concepts in physics, particularly in areas like motion and waves.