applications of trigonometry class 10 case study questions

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CBSE Class 10 Maths Case Study Questions for Chapter 9 - Some Applications of Trigonometry (Published By CBSE)

Check case study questions for cbse class 10 maths chapter 9 - some applications of trigonometry. these questions are published by the cbse itself for class 10 students..

Case study based questions are new for class 10 students. Therefore, it is quite essential that students practice with more of such questions so that they do not have problem in solving them in their Maths board exam. We have provided here the case study questions for CBSE Class 10 Maths Chapter 9 - Some Applications of Trigonometry. All these questions have been published by the Central Board of Secondary Education (CBSE) for the class 10 students. Therefore, students must solve all the questions seriously so that they may score the desired marks in their Maths exam.

Check Case Study Questions for Class 10 Maths Chapter 9:

CASE STUDY 1:

A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height.

1. What is the angle of elevation if they are standing at a distance of 42m away from the monument?

Answer: b) 45 o

2. They want to see the tower at an angle of 60 o . So, they want to know the distance where they should stand and hence find the distance.

Answer: a) 25.24 m

3. If the altitude of the Sun is at 60 o , then the height of the vertical tower that will cast a shadow of length 20 m is

a) 20√3 m

b) 20/ √3 m

c) 15/ √3 m

d) 15√3 m

Answer: a) 20√3 m

4. The ratio of the length of a rod and its shadow is 1:1. The angle of elevation of the Sun is

5. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is

a) corresponding angle

b) angle of elevation

c) angle of depression

d) complete angle

Answer: a) corresponding angle

CASE STUDY 2:

A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi(height 7,816m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of the two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

1. The distance of the satellite from the top of Nanda Devi is

a) 1139.4 km

b) 577.52 km

d) 1025.36 km

Answer: a) 1139.4 km

2. The distance of the satellite from the top of Mullayanagiri is

Answer: c) 1937 km

3. The distance of the satellite from the ground is

Answer: b) 577.52 km

4. What is the angle of elevation if a man is standing at a distance of 7816m from Nanda Devi?

5.If a mile stone very far away from, makes 45 o to the top of Mullanyangiri mountain. So, find the distance of this mile stone from the mountain.

a) 1118.327 km

b) 566.976 km

Also Check:

Case Study Questions for All Chapters of CBSE Class 10 Maths

Tips to Solve Case Study Based Questions Accurately

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Class 10 Maths: Case Study Questions of Chapter 9 Some Applications of Trigonometry PDF

Case study Questions on the Class 10 Mathematics Chapter 9  are very important to solve for your exam. Class 10 Maths Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Some Applications of Trigonometry Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 9 Some Applications of Trigonometry

Case Study/Passage Based Questions

There are two temples on each bank of a river. One temple is 50 m high. A man, who is standing on the top of a 50 m high temple, observed from the top that the angle of depression of the top and foot of other temples are 30° and 60° respectively.

The measure of ∠ADF is equal to (a) 45° (b) 60° (c) 30° (d) 90°

Answer: (c) 30°

A measure of ∠ACB is equal to (a) 45° (b) 60° (c) 30° (d) 90°

Answer: (b) 60°

Width of the river is (a) 28.90 m (b) 26.75 m (c) 25 m (d) 27 m

Answer: (a) 28.90 m

The height of the other temple is (a) 32.5 m (b) 35 m (c) 33.33 m (d) 40 m

Answer: (c) 33.33 m

The angle of depression is always (a) reflex angle (b) straight (c) an obtuse angle (d) an acute angle

Answer: (d) an acute angle

Rohit is standing at the top of the building observes a car at an angle of 30°, which is approaching the foot of the building at a uniform speed. 6 seconds later, the angle of depression of the car formed to be 60°, whose distance at that instant from the building is 25 m.

The height of the building is (a) 25√2 m (b) 50 m (c) 25√3 m (d) 25 m

Answer: (c) 25√3 m

Distance between two positions of the car is (a) 40 m (b) 50 m (c) 60 m (d) 75 m

Answer: (b) 50 m

Total time is taken by the car to reach the foot of the building from the starting point is (a) 4 secs (b) 3 secs (c) 6 secs (d) 9 secs

Answer: (d) 9 secs

The distance of the observer from the car when it makes an angle of 60° is (a) 25 m (b) 45 m (c) 50 m (d) 75 m

Answer: (c) 50 m

The angle of elevation increases (a) when point of observation moves towards the object (b) when point of observation moves away from the object (c) when object moves away from the observer (d) None of these

Answer: (a) when point of observation moves towards the object

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Maths Some Applications of Trigonometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study on Some Applications of Trigonometry Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Some Applications of Trigonometry Class 10 Maths can use this page to download the PDF file. 

The case study questions on Some Applications of Trigonometry are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Some Applications of Trigonometry case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Some Applications of Trigonometry Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Some Applications of Trigonometry, therefore, they prepared a set of solutions along with the case study questions on Some Applications of Trigonometry.

The case study on Some Applications of Trigonometry Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Some Applications of Trigonometry case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Some Applications of Trigonometry Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Some Applications of Trigonometry case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Some Applications of Trigonometry Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Some Applications of Trigonometry case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Some Applications of Trigonometry Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Some Applications of Trigonometry as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Some Applications of Trigonometry?

Students can choose their own way to answer Case Study on Some Applications of Trigonometry Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Some Applications of Trigonometry Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Some Applications of Trigonometry questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Some Applications of Trigonometry Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Some Applications of Trigonometry?

 A few essential things to know to solve Case Study Questions on Class 10 Some Applications of Trigonometry are -

• Basic Formulas of Some Applications of Trigonometry: One of the most important things to know to solve Case Study Questions on Class 10 Some Applications of Trigonometry is to learn about the basic formulas or revise them before solving the case-based questions on Some Applications of Trigonometry.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Some Applications of Trigonometry case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Some Applications of Trigonometry Class 10 Maths?

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Important Questions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry

• Class 10 Important Question
• Chapter 9: Some Applications Of Trigonometry

CBSE Class 10 Maths Important Questions Chapter 9 - Some Applications of Trigonometry - Free PDF Download

Trigonometry is always an important concept in mathematics. Although some may find it a complicated topic, once the basics are straightforward, it is very easy to tackle various levels. Moreover, practising is the best tool for any math concept, the same as trigonometry. Applications of trigonometry can be seen in many daily life cases. Furthermore, it is a very scoring topic for board exams too. Class 10 Maths Chapter 9 is an important topic from the exam point of view. Good capture of this topic will help you score good marks from the exam point of view.

Vedantu is a platform that provides free CBSE Solutions and other study materials for students. Students can register and get access to the best and reliable source of study materials specially made by master teachers at Vedantu. You can Download Maths NCERT Solutions Class 10 to help you to revise the complete Syllabus and score more marks in your examinations. Subjects like Science, Maths, Engish will become easy to study if you have access to Class 10 Science NCERT Solutions , Maths solutions, and solutions of other subjects that are available on Vedantu only.

Download CBSE Class 10 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 10 Maths Important Questions for other chapters:

CBSE Class 10 Maths Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6
7	Chapter 7
8	Chapter 8
9	Chapter 9	Some Applications of Trigonometry
10	Chapter 10
11	Chapter 11
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15

Related Chapters

Important Questions for CBSE Class 10 Maths Chapter 9 - Free PDF Download

Important questions for class 10 maths chapter 9 – Applications of Trigonometry are prepared with the vision to enable students to understand the essential topics of the chapter and study accordingly. Practising these questions can bring in-depth knowledge for students on the same subject. It is known that conceptual understanding has a pivotal role in maths. Students would be able to do questions only after they understand the concept in a clear cut manner. Moreover, practising these questions can give an excellent idea to understand which portions weigh more and know from where we have to expect questions for the board exam.

Applications of Trigonometry:

Applications of trigonometry chapter is a continuation of the previous chapter of trigonometry. Here we deal with applications of the concepts learned in the chapter trigonometry. We use the concepts of trigonometry learned during the last chapter to solve questions that include situations that we may face in our daily lives.

Suppose if you are standing near a tall building and you want to measure its height, or else suppose you are just standing on a bridge across a river and you want to measure a river's width. How can you measure the height of the building and the width of the river in both cases?

To measure the building's length, we have to go to the top of the building, drop a tall rope, and then measure the string's length. Calculating the width of the river can be more difficult as if we have to stand on both sides of the river with a long rope. These methods may seem weird for most of us as they do not seem practical. But once you are now trigonometry, the task will be easy.

Applications of trigonometry can very well be used to measure heights and distances too in an accurate manner. Trigonometry has several other applications. However, to get into the trigonometry applications, it is imperative to revise the basics of trigonometry.

Basics of Trigonometry

Trigonometric ratios form the basic step of each question from the topic. Trigonometric ratios are always derived from the sides of the right angle. Let us see some of the standard ratios used for applying trigonometry:

Sin = Perpendicular / Hypotenuse

Cos = Base / Hypotenuse

Tan = Perpendicular / Base 

Cosec = 1 / sin =Hypotenuse / Perpendicular

Sec = 1 / cos = Hypotenuse / Base

Cot = 1 / tan = Base / Perpendicular

To solve any question on heights and distances, the first and foremost thing is to draw a clean, neat, and friendly diagram labelling all the available angles and sides. The point of observation or measurement should also be included as a point in the triangle representing the question. Right angle or 90 degree is necessary to apply trigonometric ratios in such problems. The height, base, and hypotenuse should also be appropriately marked in the diagram to simplify it.

Line of Sight

Line of sight is a critical concept in trigonometry as it is based on the line of sight that angle of elevation and angle of depression is measured. When we look upon an object, an imaginary straight line connects our eyes, and the object is called the line of sight.

The angle of elevation is the angle made between the horizontal and our line of vision when we look up, and angle of depression is the angle between horizontal and the line of vision when we look down upon an object.

Besides the right angle or 90-degree angle, the angle of elevation and angle of depression plays a significant role in measuring the height of buildings or the height of any such things from an observer's point of view. Based on these angles, trigonometric ratios are applied, and the base and height are decided. If we look upon a building whose height is to be measured, the angle made is the angle of elevation, and if we are looking down for something whose depth is to be measured, the angle made is the angle of depression.

The unknown values which represent the height or distance to be measured are calculated in such a way by applying equations of trigonometric ratios to both known and unknown values. From it, the unknown is calculated. We must always take care that the unknown value should represent any one of the equation variables and apply trigonometric ratio, which includes that side. The unknown value may be based on, height or hypotenuse of the triangle. 

Practising Important Questions of Class 10 Maths Chapter 9- Applications of trigonometry are the ultimate method to tackle any kind of questions from the topic. As we say Practice makes a man perfect, the same is the case for trigonometry too. Once the basics are clear, students should go on solving NCERT questions and other important questions as well.

Practice Questions of Chapter 9

Some of the questions that can help students with their preparations for upcoming board examinations are mentioned below.

Find the height of the tower from 20m from the foot of the tower with an elevation angle of 30 degrees.

Answer: 11.56 m.

When a staircase is lying against a wall, it forms a 60° angle with the horizontal. Calculate the length of the ladder if the foot of the ladder is 2.5 metres from the wall.

Answer: 5 m

The angle of elevation of the top of a tower from a location 20 metres away is 30 degrees. Determine the tower's height.

Answer : 11.56 m

A flagstaff perches atop a 5m tall structure. The angle of elevation of the top of the flagstaff is 60 degrees from a point on earth, while the angle of elevation of the top of the structure is 45 degrees from the same point. Determine the flagstaff's height.

Answer:   3.65 m

The foot of a tower is reached along a straight highway. A man standing at the top of the tower notices a car approaching the foot of the tower at a consistent pace at a 30° angle of depression. The angle of dip of the automobile is found to be 60 degrees six seconds later. Calculate the time it took the car to go to the foot of the tower from this location.

Answer: 3 sec  

Solved Question and Answers

1. If sec 2A = cosec (A – 60°), where 4A is an acute angle, find the value of A.

Ans: A = 50°

2. Mahima is given the trigonometric ratio of tan θ = 5/12. How to find the trigonometric ratio of cosec θ using trigonometry formulas.

Ans: Using trigonometry formulas, cosec θ = 13/5

3. If sin θ cos θ = 5, find the value of (sin θ + cos θ) 2 using the trigonometry formulas.

Ans:  11

4. Find the exact value of sin 75° using the trigonometric identities.

Ans: Sin 75°= (√3 + 1)/2√2

5. Given 15 cot A = 8, find sin A and sec A

Ans: sin A = 15/17 and sec A = 17/8.

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Clarifies fundamental concepts, fostering a clear understanding.

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Simplifies applications of trigonometry, making it easier for students.

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Trigonometry is about understanding the relationships in right-angled triangles, especially the ratios of their sides called trigonometric ratios. The article above shares essential trigonometric formulas and introduces "Important Questions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry." These questions help you apply trigonometry in real situations, making it easier to grasp. Connecting theory with practical examples, this resource becomes a handy guide, ensuring you get the hang of trigonometric concepts and their use in triangles. It's like a toolkit for tackling math problems related to triangles. 

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FAQs on Important Questions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry

1. Why is Trigonometry a Difficult Topic for Many Students? Is there Any Method Through Which it Can be Learned Easily?

It is only a myth that trigonometry is a scarily difficult topic. Yes, it is true that some may find it difficult to solve the problems of the chapter, but every chapter may have such issues. But success lies where we do not stop trying more and more unless we become experts on the topic. As we the way to success do not have any shortcuts, the same is the trigonometry case. We have to first be thorough with all the trigonometric ratios and equations and then learn to draw diagrams in the most accurate manner and then keep on practising and practising. Our team's important questions at Vedantu can help you a lot in this process. We can assure you that none of your topics is left uncovered once you have done all the questions prepared by us. Practice questions to the maximum so that you become able to solve problems at lightning speed. Practising more is equal to scoring more and more. So, once the basics are clear, collect questions from whatever sources possible and start solving them at the earliest.

2. Is Trigonometry and Applications of Trigonometry Merely a Piece for Academic Scoring? Or Whether it Has Any Purpose in Our Daily Life?

Actually, there is a false belief that there is something merely for academics. Everything that we learn in our books will surely have some applications in our daily life. The truth is that we are just not aware of it. The same is the case for trigonometry and also for applications of trigonometry. From the name of the chapter – Applications of trigonometry itself, we can understand that it is an application-oriented chapter. In many cases, such as finding length, depth, distance, etc., were measured by the manual process can turn into a very hectic process, applications of trigonometry have helped us. Instead of using many instruments for measuring, with just a pen and paper and some trigonometric ratios, we can do the same. Trigonometry finds wide applications in the field of architecture, engineering, and many such. Besides in daily life, when coming to academics also, trigonometry is a very scoring chapter that can be easily dealt with if we have practised enough. As every bookish knowledge is meant to be applied in our life, the same is the case for trigonometry too. 

3. What is the significance of Chapter 9 “Some Applications of Geometry” of Class 10 Maths?

Chapter 9 “Some Applications of Geometry” of Class 10 Maths has much significance both from the examination perspective as well as due to its applications in our daily lives. 

For exams, this chapter is important because i) it can carry around 6 marks ii) it is a chapter that can help you score high in exams.

Moreover, this concept has several applications in fields like Engineering, Criminology, Marine Biology, Constructions, Aeronautics, Navigation, Physics, measuring the heights of buildings, mountains, etc. Hence, learning it will be beneficial. 

4. How can I download Vedantu’s Important Questions for Chapter 9 “Some Applications of Geometry” of Class 10 Maths.

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5. What are the significant features of Vedantu’s Important Questions for Chapter 9 “Some Applications of Geometry” of Class 10 Maths?

These are the very advantageous features of Vedantu's Important Questions:

These are cautiously selected by expert teachers.

These have been put together after screening several sample papers, mock papers, reference books, and previous years' question papers.

These have been systematically arranged as per the marking scheme of 1 to 5 marks questions.

 The solutions to these questions are prepared especially to focus on the CBSE board examinations.

These are available to be downloaded free of charge.

6. What should I do if I find Chapter 9 “Some Applications of Geometry” of Class 10 Maths difficult?

It is perfectly normal for students to find certain chapters intimidating. However, the first step to get over this is to let go of the fear. Analyze why you find the chapter complicated and work on the problem areas in parts.

Read the text thoroughly and refer to visual aids to help in enhanced understanding. Ask your teachers in case of doubts. You can also opt for Vedantu's one-on-one classes for the same. Practising a good number of problems will help you overcome any fear.

7. What are the three main topics taught in Chapter 9 “Some Applications of Geometry” of Class 10 Maths?

The three major topics that help in understanding various applications of trigonometry are as follows:

Line of sight: refers to a straight line from the eye of the observer to the object

Angle of elevation: the angle formed by the line of sight and the horizontal line. This happens in case the observer is looking upwards at the object.

Angle of depression: the angle formed by the line of sight and the horizontal line. This is used when the observer is looking downwards at the object.

CBSE Class 10 Maths Important Questions

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Class 10 Math Chapter 9 Applications of Trigonometry Important Questions

Table of Contents

Introduction

What is the angle of elevation and angle of depression in trigonometry.

Class 10 Applications of Trigonometry Important Questions and Answers

Q1. a tower stands vertically on the ground. from a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45°. then the height (in meters) of the tower is :, (a) 25\sqrt{2}\\ (b) 25\sqrt{3}\\ (c) 25 (d) 12.5.

Ans. (c) 25 Explanation: Let the height of the tower be H m.

Q2. The angle of depression of a car standing on the ground from the top of a 75 m high tower is 30°. The distance of the car from the base of the tower (in m) is :

(a) 25 \sqrt{3}\\ (b) 50 \sqrt{3}\\ (c) 75 \sqrt{3}\\ (d) 150.

Ans. (c) 75 \sqrt{3}\\ Explanation: Let the distance of the car from the base of the tower be x m.

Q3. From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°. Find the (i) height of the tower, (ii) the depth of the tank.

Ans. The height of the tower is h = 23.1 m and the depth of the tank is h= 16.9 m Explanation: Let BC be the tower of height h m and CD be the water tank of height h1 m. Let A be a point on the ground at a distance of 40 m away from the foot B of the tower.

Q4. The angle of elevation of an airplane from a point on the ground is 60°.After a flight of 30 seconds, the angle of elevation becomes 30°. If the airplane is flying at a constant height of 3000

Ans. The speed of the airplane is 720 km/hr. Explanation: Let the ground distance between the airplane A and the point E be x m.

Given, height AD is 3000 \sqrt{3} \text{m } and the angle of elevation is 60° So, in ∆AED, \text{tan 60}^o = \dfrac{3000\sqrt{3}}{x} \\ ⇒ \sqrt{3} = \dfrac{3000\sqrt{3}}{x}\\ \text{⇒ x = 3000 m }\\ Let the new distance (DC) covered by the airplane in 30 seconds be y m. \\ \text{So, tan 30}^o = \dfrac{3000\sqrt{3}}{3000 + y} \\[4.5 bp] ⇒ \dfrac{1}{\sqrt{3}}= \dfrac{3000\sqrt{3}}{3000 + y}\\[4.5 bp] ⇒ 3000 + y = (3000\sqrt{3})\sqrt{3}\\[4.5 bp] ⇒ 3000 + y = 9000\\[4.5 bp] \text{⇒ y = 9000 - 3000 = 6000 m}\\[4.5 bp] Thus, distance covered in 30 seconds = 6000 m \\ \text{Hence, Speed = } \dfrac{6000}{30} \\ \text{= 200 m/sec}\\ = 200 × \dfrac{18}{5}\text{km/hr.}\\ Hence, the speed of the airplane is 720 km/hr.

Q5. From the top of a tower, the angles of depression of two objects on either side of the tower are found to be α and β(α < β). If the distance between them is d m, show that the height of the tower is given by

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Q1 : how can application of trigonometry be applied to find the height of a building or a tree, q2 : how can trigonometry be applied to find the distance between two objects, q3 : how is trigonometry applied in solving real-life problems related to heights, q4 : in what practical situations is trigonometry applied in navigation, q5: explain the application of trigonometry in construction and engineering..

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Class 10 Maths Case Study Questions Chapter 8 Introduction to Trigonometry

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Case study Questions in the Class 10 Mathematics Chapter 8  are very important to solve for your exam. Class 10 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 8  Introduction to Trigonometry

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Introduction to Trigonometry Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Case Study/Passage-Based Questions

Question 1:

(a) 2m

(b) 3m

(c) 4m

(d) 6m

Answer: (d) 6m

(ii) Measure of ∠A =

(a) 30°

(b) 60°

(c) 45°

(d) None of these

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

(a) 0

(b) 1

(c) 1/2

(d) 2√2

Answer: (d) 2√2

(v) Find the value of tan 2 C + tan 2  A.

(a) 0

(b) 1

(c) 2

(d) 1/2

Answer: (c) 2

Question 2:

(a) 30°

(b) 45°

(c) 60°

(d) None of these

Answer: (a) 30°

(ii) The measure of  ∠C is

Answer: (c) 60°

(iii) The length of AC is 

(a)2√3 m

(b)√3m

(c)4√3m

(d)6√3m

Answer: (d)6√3m

(iv) cos2A =

(a) 0

(b)1/2

(c)1/√2

(d)√3/2

Answer: (b)1/2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Introduction to Trigonometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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• NCERT Solutions
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• NCERT Solutions for Class 10 Maths
• Chapter 9: Some Applications Of Trigonometry

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry

Ncert solutions for class 10 maths chapter 9 – cbse get free pdf.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry provides comprehensive solutions for all the questions in the NCERT   Textbook . To excel in the CBSE board examinations, NCERT Solutions will increase the level of confidence among the students, as the concepts are clearly explained and structured. The solutions are prepared and reviewed by our subject matter experts, and are revised according to the latest CBSE Syllabus for 2023-24 and guidelines of the CBSE board.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 9 Some Applications of Trigonometry

Download most important questions for class 10 maths chapter – 9 some applications of trigonometry.

It covers all the chapters and provides chapter-wise solutions. These NCERT Solutions for Class 10 Maths  are helpful for the students to clarify their doubts and provide a strong foundation for every concept. With the help of NCERT Solutions for Class 10 , every student will be capable of easily solving the complex problem in each exercise.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry…
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
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Access Answers to NCERT Class 10 Maths Chapter 9 – Some Applications of Trigonometry

Exercise 9.1 page no: 203.

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. 9.11)

Length of the rope is 20 m and angle made by the rope with the ground level is 30°.

Given: AC = 20 m and angle C = 30°

To Find : Height of the pole

Let AB be the vertical pole

In right ΔABC, using sine formula

sin 30° = AB/AC

Using value of sin 30 degrees is ½, we have

1/2 = AB/20

Therefore, the height of the pole is 10 m.

2. 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.

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°

To Find: Height of the tree, which is AB

From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ΔABC,

Using Cosine and tangent angles,

cos 30° = BC/AC

We know that, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.

3. 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?

As per contractor’s plan,

Let, ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at

60° with length PR.

To Find: AC and PR

1/2 = 1.5/AC

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

Hence, length of the slide for below 5 = 3 m and

Length of the slide for elders children = 2√3 m

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.

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

To Find: AB (height of the tower)

In right ABC

1/√3 = AB/30

⇒ AB = 10√3

Thus, the 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.

Draw a figure, based on given instruction,

Let BC = Height of the kite from the ground, BC = 60 m

AC = Inclined length of the string from the ground and

A is the point where string of the kite is tied.

To Find: Length of the string from the ground i.e. the value of AC

From the above figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Thus, the length of the string from the ground is 40√3 m.

6. 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.

Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

To Find: The distance boy walked towards the building i.e. XY

From figure,

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

tan 60° = AB/BC

√3 = 28.5/BC

BC = 28.5/√3 = 28.5√3/3

Therefore, the length of BC is 28.5√3/3 m.

XY = CD = BD – BC = (28.5√3-28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 = 19√3  m.

Thus, the distance boy walked towards the building is 19√3 m.

7. 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.

Let BC be the 20 m high building.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC – BC

To Find: AB, Height of the tower

From figure, In right ΔBCD,

tan 45° = BC/CD

In right ΔACD,

tan 60° = AC/CD

Now, AB = AC – BC = (20√3-20) = 20(√3-1)

Height of transmission tower = 20(√3 – 1) m.

8. 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.

Let AB be the height of statue.

To Find: Height of pedestal = BC = AC-AB

In right triangle BCD,

BC = CD …..(1)

√3 = ( AB+BC)/CD

√3CD = 1.6 + BC

√3BC = 1.6 + BC (using equation (1)

√3BC – BC = 1.6

BC(√3-1) = 1.6

BC = [1.6(√3+1)]/(2) m

BC = 0.8(√3+1)

Thus, the height of the pedestal is 0.8(√3+1) m.

9. 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.

Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.

In right ΔBCD,

tan 60° = CD/BC

BC = 50/√3 …(1)

⇒ 1/√3 = AB/BC

Use result obtained in equation (1)

Thus, the height of the building is 50/3 m.

10. 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.

Let AB and CD be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

As per above figure, AB = CD,

OB + OD = 80 m

In right ΔCDO,

tan 30° = CD/OD

1/√3 = CD/OD

CD = OD/√3 … (1)

In right ΔABO,

tan 60° = AB/OB

√3 = AB/(80-OD)

AB = √3(80-OD)

AB = CD (Given)

√3(80-OD) = OD/√3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

Putting the value of OD in equation (1)

CD = 20√3 m

⇒ OB = (80-60) m = 20 m

Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and

60 m respectively.

11. 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 joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Solution: Given, AB is the height of the tower.

DC = 20 m (given)

As per given diagram, In right ΔABD,

1/√3 = AB/(20+BC)

AB = (20+BC)/√3 … (i)

AB = √3 BC … (ii)

From equation (i) and (ii)

√3 BC = (20+BC)/√3

3 BC = 20 + BC

Putting the value of BC in equation (ii)

This implies, the height of the tower is 10√3 m and the width of the canal is 10 m.

12. 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.

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

Design a figure based on given instructions:

tan 45° = AB/BC

Since BC = AD

Again, from right triangle ADE,

tan 60° = DE/AD

⇒ DE = 7√3 m

Now: EC = DE + CD

= (7√3 + 7) = 7(√3+1)

Therefore, height of the tower is 7(√3+1) m. Answer!

13. 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.

Let AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.

30° and 45° are the angles of depression from the lighthouse.

Draw a figure based on given instructions:

To Find: CD = distance between two ships

Step 1: From right triangle ABC,

Step 2: Form right triangle ABD,

1/√3 = 75/BD

Step 3: To find measure of CD, use results obtained in step 1 and step 2.

CD = BD – BC = (75√3 – 75) = 75(√3-1)

The distance between the two ships is 75(√3-1) m. Answer!

14. 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 Fig. 9.13). Find the distance travelled by the balloon during the interval.

Let the initial position of the balloon be A and final position be B.

Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m.

To Find: Distance travelled by the balloon = DE = CE – CD

Let us redesign the given figure as per our convenient

Step 1: In right ΔBEC,

tan 30° = BE/CE

1/√3= 87/CE

In right ΔADC,

tan 60° = AD/CD

CD = 87/√3 = 29√3

DE = CE – CD = (87√3 – 29√3) = 29√3(3 – 1) = 58√3

Distance travelled by the balloon = 58√3 m.

15. 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.

Let AB be the tower.

D is the initial and C is the final position of the car respectively.

Since the man is standing at the top of the tower so, Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

Step 1: In right ΔABC,

1/√3 = AB/BD

Step 3: Form step 1 and Step 2, we have

√3 BC = BD/√3 (Since LHS are same, so RHS are also same)

3 BC = BC + CD

or BC = CD/2

Here, distance of BC is half of CD. Thus, the time taken is also half.

Time taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6/2 = 3 sec.

16. 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.

Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per question,

tan x = AB/BC

tan x = AB/4

AB = 4 tan x … (i)

Again, from right ΔABD,

tan (90°-x) = AB/BD

cot x = AB/9

AB = 9 cot x … (ii)

Multiplying equation (i) and (ii)

AB 2 = 9 cot x × 4 tan x

⇒ AB 2 = 36 (because cot x = 1/tan x

Since height cannot be negative, therefore, the height of the tower is 6 m.

Hence Proved.

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

For the  Class 10 CBSE Maths paper, out of 80 marks, 12 marks are assigned from the chapter “ Trigonometry”. You can expect 1 question from this chapter. The paper consists of 4 parts and each carries different marks. The questions have been assigned with 1 mark, two marks, 3 marks and 4 marks.

The main topics covered in Class 10 NCERT Maths of this chapter include:

9.1 introduction.

In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying some ways in which trigonometry is used in the life around you. Trigonometry is one of the most ancient subjects studied by scholars all over the world. As we have studied in Chapter 8, trigonometry was invented because its need was identified in astronomy. In this chapter, we will see how trigonometry is used for finding the heights and distances of various objects, without actually measuring them.

9.2 Heights and Distances

The topic discusses the line of sight, angle of deviation, angle of elevation and angle of depression. All the processes are explained by solving some problems. The numerical problems are solved with the help of trigonometric ratios.

9.3 Summary

The summary describes all the points you have studied in the chapter. It will help you to understand the important concepts that need to be focused upon from the chapter.

Students can utilise the NCERT Solutions for Class 10 to attain a firm grip over the key concepts present in Class 10 Maths Chapter 9. These solutions are prepared by well-experienced teachers at BYJU’S with the aim to render clarity on key concepts and problem-solving skills.

List of Exercises in Class 10 Maths Chapter 9:

Exercise 9.1 Solutions – 16 Questions (16 long answers)

In this chapter, “ Some applications of trigonometry ”, Class 10 students will get to know how trigonometry is helpful in finding the height and distance of different objects without measuring. In earlier days, astronomers have used trigonometry for calculating the distance from the earth to the planets and stars. Trigonometry is mostly used in navigation and geography to locate the position in relation to the latitude and longitude.

Students will learn the applications of trigonometry with real-life examples in a better way. With the help of geometrical figures, the important terms and problems are explained and the summary is given at the end of the chapter. In NCERT Solutions for Class 10 Maths , you are provided with step-by-step procedure solutions to all the questions.

Key Features of NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry

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The expert faculty team of members have formulated the NCERT Solutions in a lucid manner to improve the problem-solving abilities among the students. For a more clear idea about some applications of trigonometry, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Some applications of trigonometry

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Dropped Topics –  9.1 Introduction

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CBSE Case Study Questions for Class 10 Maths Trigonometry Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Trigonometry  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Trigonometry PDF

Checkout our case study questions for other chapters.

• Chapter 6 Triangles Case Study Questions
• Chapter 7 Coordinate Geometry Case Study Questions
• Chapter 9 Some Applications of Trigonometry Case Study Questions
• Chapter 10 Circles Case Study Questions

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Class 10th Maths - Some Applications of Trigonometry Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Some Applications of Trigonometry, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Some applications of trigonometry case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

Case Study 

(ii) Measure of \(\angle\) ACB is equal to

(iii) Width of the river is 

(iv) Height of the other temple is

(v) Angle of depression is always

(ii) Value of DF is equal to

\((a) \frac{h}{\sqrt{3}} \mathrm{~m}\)

\((b) h \sqrt{3} \mathrm{~m}\)

\((c) \frac{h}{2} \mathrm{~m}\)

\((d) 2 h \mathrm{~m}\)

(iii) Value of h is

(iv) Height of the balloon from the ground is

(v) If the balloon is moving towards the building, then both angle of elevation will

\((a) 15 \mathrm{~m}\)	\((b) 15 \sqrt{2} \mathrm{~m}\)
\((c) \frac{15}{\sqrt{3}} \mathrm{~m}\)	\((d) \frac{15}{\sqrt{2}} \mathrm{~m}\)

(ii) If the angle made by the rope to the ground level is 45°, then find the distance between artist and pole at ground level.

\((a) \frac{15}{\sqrt{2}} \mathrm{~m}\)

\((b) 15 \sqrt{2} \mathrm{~m}\)

\((c) 15 \mathrm{~m}\)

\((d) {15}{\sqrt{3}} \mathrm{~m}\)

(iii) Find the height of the pole if the angle made by the rope to the ground level is 30°.

(iv) If the angle made by the rope to the ground level is 30° and 3 m rope is broken, then find the height of the pole

(v) Which mathematical concept is used here?

(ii) If fireman place the ladder 5 m away from the wall and angle of elevation is observed to be 30°, then length of the ladder is

\((b) \frac{10}{\sqrt{3}} \mathrm{~m}\)

\((c) \frac{15}{\sqrt{2}} \mathrm{~m}\)

(iii) If fireman places the ladder 2.5 m away from the wall and angle of elevation is observed to be 60°, then find the height of the window. (Take \(\sqrt{3}\) = 1.73)

(iv) If the height of the window is 8 m above the ground and angle of elevation is observed to be 45°, then horizontal distance between the foot of ladder and wall is

(v) If the fireman gets a 9 m long ladder and window is at 6 m height, then how far should the ladder be placed?

\(\sqrt{5}\)

(ii) What should be the length of ladder, so that it makes an angle of 60° with the ground?

\((a) 4\sqrt{3} {~m}\)

\((b) 2\sqrt{3} {~m}\)

\((c) 3\sqrt{3} {~m}\)

\((d) 5\sqrt{3} {~m}\)

(iii) The distance between the foot ofladder and pole is

\((a) 6\sqrt{3} {~m}\)

\((b) 4\sqrt{3} {~m}\)

\((c) 3\sqrt{3} {~m}\)

\((d) 2\sqrt{3} {~m}\)

(iv) What will be the measure of \(\angle\) BCD when BD and CD are equal?

(v) Find the measure of \(\angle\) DBC.

(ii) Distance between two positions of the car is

(iii) Total time taken by the car to reach the foot of the building from starting point is

(iv) The distance of the observer from the car when it makes an angle of 60° is

(v) The angle of elevation increases

(ii) If  \(\angle\) YAB = 30°, then \(\angle\) ABD is also 30°, Why?

(iii) Length of CD is equal to

(iv) Length of BD is equal to

m

m

(v) Length of AC is equal to

m

m

(ii) If the height of the pedestal is 20 m, then the height of the statue is

\((a) 20 \sqrt{3} \mathrm{~m}\)

\((b) 20(\sqrt{3}-1) \mathrm{m}\)

\((c) 20(\sqrt{3}+1) \mathrm{m}\)

\((d) 10(\sqrt{3}-1) \mathrm{m}\)

(iii) If the height of the statue is 1.6 m, then height of the pedestal is

\((a) 0.8(\sqrt{3}-1) \mathrm{m}\)

\((b) 1.6(\sqrt{3}+1) \mathrm{m}\)

\((c) 0.8(\sqrt{3}) \mathrm{m}\)

\((d) 0.8(\sqrt{3}+1) \mathrm{m}\)

(iv) If the total height of the statue and pedestal is 39 m, then find the length of AC.

(v) If the height of the pedestal is 35 m, then length of AD is

\((a) (\angle x, \angle y)\)

\((b) (\angle y, \angle z)\)

\((c) (\angle z, \angle t)\)

\((d) (\angle r, \angle q)\)

(ii) If the position of Pankaj is 25 m away from the base of pedestal and Zr = 30°, then find the height of pedestal.

(iii) If the height of pedestal is 30 m, \(\angle\) t = 45° and \(\angle\) z = 30°, then the horizontal distance between Arun and Pankaj is

(iv) If the vertical height of sky lantern from the top of pedestal is 12 m and \(\angle\) y = 30°, then distance between Teewan and sky lantern is

(v) If \(\angle\) q = 60° and position of Arun is 15 m away from the base of pedestal, then find the height of pedestal.

(ii) Find the length of RO.

(iii) The width of the road is

(iv) If the angle of elevation made by pole PQ is 45°, then the length of PO =

(v) Angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level is known as

\((a) \sqrt{3} \mathrm{~m}\)

\((b) 30 \sqrt{3} \mathrm{~m}\)

\((c) \frac{30}{\sqrt{3}} \mathrm{~m}\)

(ii) If the top of broken part of a tree touches the ground at a point whose distance from foot of the tree is equal to height of remaining part, then its angle of inclination is

(iii) The angle of elevation are always

(v) If the height of a tree is 6 m, which is broken by wind in such a way that its top touches the ground and makes an angles 30° with the ground. At what height from the bottom of the tree is broken by the wind?

\((b) \frac{\sqrt{3}}{6} \mathrm{~m}\)

\((c) 6 \sqrt{3} \mathrm{~m}\)

\((d) \frac{6}{\sqrt{3}} \mathrm{~m}\)

(a) 6 m

\((b) 6 \sqrt{3} \mathrm{~m}\)

\((c) \sqrt{3} \mathrm{~m}\)

\((d) 10 \sqrt{3} \mathrm{~m}\)

(iii) Width of the river is

(iv) The angles of elevation and depression are always

(v) If BD = 21 m, then height of the bridge is

\((c) 7 \sqrt{3} \mathrm{~m}\)

\((d) \frac{7}{\sqrt{3}} \mathrm{~m}\)

(ii) The value of PD is

(v) If A and B are two objects and the eye of an observer is at point 0, then the line of sight will be

\((c) \frac{100}{\sqrt{3}} \mathrm{~m}\)

\((d) 100 \sqrt{3}\)

(ii) If the distance between the position of pigeon increases, then the angle of elevation

(iii) Find the distance between the boy and the pole.

\((b) \frac{50}{\sqrt{3}} \mathrm{~m}\)

\((c) 50 \sqrt{3} \mathrm{~m}\)

\((d) 60 \sqrt{3} \mathrm{~m}\)

(iv) How much distance the pigeon covers in 8 seconds?

(v) Find the speed of the pigeon

(ii) Find the length of GH.

(iii) The length of second step is 

(iv) The length of PQ =

(v) The length of first step is

*****************************************

Some applications of trigonometry case study questions with answer key answer keys.

(i) (b): The person who makes small angle of elevation is more closer to the balloon. \(\therefore\) Radlra is more closer to the balloon. (ii) (b):  \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\) \(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \) \(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\) (iii) (a): In \(\Delta\) GCE, \(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\) (iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m (v) (b)

(i) (b): Total height of pole = 8 m \(\therefore\) BD = AD - AB = (8 - 2)m = 6 m (ii) (a):  \(\text { In } \Delta B D C, \frac{B D}{B C}=\sin 60^{\circ}\) \(\Rightarrow \quad \frac{6}{B C}=\frac{\sqrt{3}}{2} \) \(\Rightarrow \quad B C=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\) (iii) (d):  \(\text { In } \triangle B D C\) \(\frac{B D}{C D}=\tan 60^{\circ} \Rightarrow \frac{6}{C D}=\sqrt{3} \Rightarrow C D=\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\) (iv) (b) :  \(\text { If } \Delta B C D\) \(\frac{B D}{C D}=\tan \theta \Rightarrow 1=\tan \theta \quad[\because B D=C D] \) \(\Rightarrow \quad \theta=45^{\circ}\) (v) (c) :   \(\operatorname{In} \Delta B D C, \angle B+\angle D+\angle C=180^{\circ}\) \(\therefore \quad \angle B=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}\)

(i) (c):   \(\text { In } \Delta A B C, \frac{A B}{B C}=\tan 60^{\circ}\) \(\Rightarrow \quad A B=25 \times \sqrt{3}\) \(\therefore\) Height of building is 25 \(\sqrt{3}\) m . (ii) (b):   \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 30^{\circ}\) \(\Rightarrow \frac{25 \sqrt{3}}{B D}=\frac{1}{\sqrt{3}} \Rightarrow B D=75 \mathrm{~m}\) \(\therefore\)  Distance between two positions of car = (75 - 25) m = 50m. (iii) (d): Time taken to cover 50 m distance = 6 sec. \(\therefore\) Time taken to cover 25 m distance = 3 sec. \(\therefore\) Total time taken by car = 6 sec + 3 sec = 9 sec (iv) (c):  \(\text { In } \Delta A B C, \frac{B C}{A C}=\cos 60^{\circ}\) \(\Rightarrow \quad \frac{25}{A C}=\frac{1}{2} \) \(\Rightarrow A C=50 \mathrm{~m}\) (v) (a)

(i) (b):   \(\angle X A C=45^{\circ}\)   \(\therefore \quad \angle A C D=45^{\circ}\)   [Alternate interior angles] (ii) (b) (iii) (c) :  \(\text { In } \Delta A C D\) \(\frac{A D}{D C}=\tan 45^{\circ} \) \(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\) (iv) (d):   \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\) \(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \) \(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\) (v) (a):  \(\text { In } \Delta A D C\) \(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

(i) (c):   \(\text { In } \Delta O P Q\) ,  we have \(\tan 60^{\circ}=\frac{P Q}{P O} \) \(\Rightarrow \sqrt{3}=\frac{20}{P O} \) \(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\) (ii) (b): In \(\Delta\) ORS, we have \(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\) (iii) (d): Clearly, width of the road = PR \(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\) (iv) (a):   \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\) \(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\) (v) (b)

(i) (d) : Clearly, \(\angle\) DAC = 60° So,in \(\Delta\) ADC, we have \(\tan 60^{\circ}=\frac{C D}{A D} \Rightarrow \sqrt{3}=\frac{6}{A D} \) \(\Rightarrow A D=\frac{6}{\sqrt{3}} \mathrm{~m}\) (ii) (b): Clearly, \(\angle\) DBC = 30° So, in  \(\Delta\) BDC,we have \(\tan 30^{\circ}=\frac{C D}{B D} \) \(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{6}{B D} \) \(\Rightarrow B D=6 \sqrt{3} \mathrm{~m}\) (iii) (b): Width of the river = AB = AD + BD \(\begin{array}{l} =\frac{6}{\sqrt{3}}+6 \sqrt{3} \\ =6\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)=6\left(\frac{4}{\sqrt{3}}\right)=\frac{24}{\sqrt{3}} \mathrm{~m}=13.87 \mathrm{~m} \end{array}\) (iv) (a): The angle of elevation and angle of depression are always acute angles. (v) (c): In \(\Delta\) BCD,if BD = 21m, then \( \tan 30^{\circ}=\frac{C D}{B D} \) \(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{C D}{21} \Rightarrow C D=\frac{21 \sqrt{3}}{3}=7 \sqrt{3} \mathrm{~m}\)

(i) (b): In the right   \(\Delta\) ADQ, we have \(\sin 30^{\circ}=\frac{D Q}{A D} \Rightarrow \frac{1}{2}=\frac{D Q}{24}\) \(\Rightarrow \quad D Q=12 \mathrm{~m}\) Thus, distance of paraglider from the ground is 12 m. (ii) (a): We have PQ = BC = 6 m Now, as DQ = 12 m \(\therefore\) DP = DQ - PQ = 12 - 6 = 6 m (iii) (c) : In right  \(\Delta\) BDP,we have \(\sin 45^{\circ}=\frac{D P}{B D} \Rightarrow \frac{1}{\sqrt{2}}=\frac{6}{B D}\) \(\Rightarrow \quad B D=6 \sqrt{2} \mathrm{~m}\) Thus, the distance of paraglider from the girl is 6 \(\sqrt{2}\)  m. (iv) (d): \(\angle\) AOP given in figure, is the angle of depression. (v) (c): If A and B are two objects and the eye of an observer is at point 0, then line of sight will be both OA and OB.

Given, side of square top = 2 m \(\therefore\) AB = HT = QR = CD = 2 m Also, A C and BD are perpendicular to the ground. So, AH = HQ = QC.  (By B.P.T. Theorem) (i) (b): \(\text { In } \triangle A E C\) \(\sin 60^{\circ}=\frac{A C}{A E} \Rightarrow \frac{\sqrt{3}}{2}=\frac{6}{A E} \Rightarrow A E=6.93 \mathrm{~m}\) \(\therefore\) Length of each leg i.e., AE = BF = 6.93 m. (ii) (c):   \(\text { In } \Delta A G H, \tan 60^{\circ}=\frac{A H}{G H} \Rightarrow \sqrt{3}=\frac{2}{G H}\) \(\Rightarrow G H=1.15 \mathrm{~m}\) (iii) (a) : Length of second step = GH + HT + TU = 1.15 + 2 + 1.15 = 4.3 m (iv) (b): \(\text { In } \Delta A P Q\) \(\tan 60^{\circ}=\frac{A Q}{P Q} \Rightarrow \sqrt{3}=\frac{4}{P Q} \Rightarrow P Q=\frac{4}{\sqrt{3}} \mathrm{~m}=2.31 \mathrm{~m}\) (v) (c) : Length of first step = PQ + QR + RS =2.31 + 2 + 2.31 = 6.62 m

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Case Based Questions: Some Application of Trigonometry - Class 10 MCQ

15 questions mcq test - case based questions: some application of trigonometry, read the following text and answer the following questions on the basis of the same. a straight highway leads to the foot of 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°. q. write the value of sec 30°..

Read the following text and answer the following questions on the basis of the same. A straight highway leads to the foot of 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°. Q. The line drawn from the eye of an observer to the point in the object viewed by the observer.

• A. horizontal line
• B. Vertical line
• C. Line of sight
• D. Parallel lines

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Read the following text and answer the following questions on the basis of the same. A straight highway leads to the foot of 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°. Q. Find the time taken by the car to reach the foot of the tower from point D to B.

Let the speed of car be v m/s.

Let car takes t seconds to reach the point B from the point D

Distance travel by car in t sec = vt m.

In ΔABD, we have

h = √3 vt ...(i)

and in right D ABC, we have

Read the following text and answer the following questions on the basis of the same.

A straight highway leads to the foot of 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°.

Q. Write the value of cosec 60°.

Q. If the two lines are parallel; then the alternate opposite angles are ..................... .

• A. different
• C. opposite
• D. None of these

Form a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and angle of elevation of the top of the flagstaff from P is 45°.

Q. What is the value of tan 45°?

AP = 10 √3 m

In right ΔPAD,

10 √3 = 10 + BD

BD = 10 √3 – 10

BD = 7.32 m.

• A. BP 2 = AB 2 + AP 2
• B. AB 2 = AP 2 + BP 2
• C. AP 2 = AB 2 + BP 2
• D. None of these.

Read the following text and answer the following questions on the basis of same.

From a point on the bridge across a river the angle of depression of the banks on opposite sides of the river 30° and 45° respectively.

• A. Acute angled triangle
• B. Right angled triangle
• C. Obtuse angled triangle
• D. Equilateral triangle.

From a point on the bridge across a river the angle of depression of the banks on opposite sides of the river 30° and 45° respectively.

Q. The value of tan 45° is

The value of tan 45° is = 1

• A. 1(√3 + 1) m
• B. ( √3 +1) m
• C. ( √3 +2) m
• D. 3(√3 + 1) m

In ΔPDB, ∠B = 45°

tan 45° = PD/DB

width of the river = AB = AD + DB

= 3(√3 + 1)m.

• A. Perpendicular/Base
• B. Base/Perpendicular
• C. Hypotenuse/Base
• D. Perpendicular/Hypotenuse

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Important Questions for Case Based Questions: Some Application of Trigonometry

Case based questions: some application of trigonometry mcqs with answers, online tests for case based questions: some application of trigonometry.

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Question 13 - Case Study - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [Term 2] - Solutions of Sample Papers for Class 10 Boards

Last updated at April 16, 2024 by Teachoo

Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites. A guard, stationed at the top of a 240m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used fo measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.(Lighthouse of Mumbai Harbour. Picture credits - Times of India Travel) i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower.

Ii) after 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(√3 - 1) m. he immediately raised the alarm. what was the new angle of depression of the boat from the top of the observation tower.

This question is similar to Ex 9.1, 13 Chapter 9 Class 10 - Some Applications of Trigonometry

Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites. A guard, stationed at the top of a 240m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used fo measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°. (Lighthouse of Mumbai Harbour. Picture credits - Times of India Travel) i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower. ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(√3 - 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower? Making a labelled figure Given that height of the lighthouse is 240 m Hence, AC = 240 m And angle of depression of boat is 30° So, ∠ PAB = 30 ° Since Angle of depression = Angle of elevation ∴ ∠ ABC = 30° Question 13 (i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower. We need to find distance between boat and tower, i.e. BC In right angled triangle ΔABC, tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan 30° = 𝐴𝐶/𝐵𝐶 (" " 1)/√3 = (" " 240)/𝐵𝐶 BC = 240√𝟑 m Question 13 (ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(√3−1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower? Let Boat be now at point D Since Distance of tower is reduced by 240(√3−1) m Hence, BD = 𝟐𝟒𝟎(√𝟑−𝟏) m Let angle of depression of boat now be θ So, ∠ PAD = θ ° Since Angle of depression = Angle of elevation ∴ ∠ ADC = θ Also, CD = BC − BD = 240√3 −240(√3−1) = 240√3 −240√3+240 = 𝟐𝟒𝟎 m Now, In right angled triangle ΔABC, tan D = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan θ = 𝐴𝐶/𝐶𝐷 tan θ = 𝟐𝟒𝟎/𝟐𝟒𝟎 tan θ = 1 ∴ θ = 45° Thus, required angle of depression is 45°

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9 with Solutions

January 7, 2023 by Sastry CBSE

Extra Questions for Class 10 Maths Applications of Trigonometry with Answers

Extra Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry. According to new CBSE Exam Pattern,  MCQ Questions for Class 10 Maths  Carries 20 Marks.

You can also download NCERT Class 10 Maths Solutions to help you to revise complete syllabus and score more marks in your examinations.

Applications of Trigonometry Class 10 Extra Questions Very Short Answer Type

Applications of Trigonometry Class 10 Extra Questions Short Answer Type-2

Applications of Trigonometry Class 10 Extra Questions Long Answer Type 1

Applications of Trigonometry Class 10 Extra Questions HOTS

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1. If the ratio of length of the shadow of a pole to its height is √3 : 1, then elevation of the sun is (a) 30° (b) 45° (c) 60° (d) 120° Answer: (c) 60°

Question 2. A steel wire is tied to the top of an electric pole and the ground making an angle of 60° with the ground. If the height of electric pole is 12 m, then length of steel wire is (a) 4√3 m (b) 8√3 m (c) \(\frac{4}{\sqrt{3}}\) m (d) 12√3 m Answer: (b) 8√3 m

Question 4. Height of a tower is 10 m. If the sun’s altitude is 45°, then length of the shadow is : (a) 5 m (b) 20 m (c) 10√2 m (d) 10 m Answer: (d) 10 m

Question 6. A tree is broken by the wind, its top struck the ground at an angle of 30° at a distance of 30 m from its foot. The whole height of tree is (a) 30 √3 m (b) 20 √3 m (c) 10√3 m (d) 40√3 m Answer: (a) 30 √3 m

Question 7. The angle of elevation of a tower from two points distant l and m (l > m) from its foot and in the same straight line from it are complementary, then the height of the tower is (a) \(\sqrt{\frac{l}{m}}\) (b) 2√lm (c) √lm (d) \(\sqrt{\frac{m}{l}}\) Answer: (c) √lm

Question 8. An observer 1.5 m tall is 28.5 m away from a tower of height 30 m. The angle of elevation of the top of tower from his eye is (a) 30° (b) 45° (c) 60° (d) 75° Answer: (b) 45°

Question 9. If the shadow of a tower standing on a level plane is found to be 50 m longer when sun’s elevation is 30° than when it is 60°, then the height of the tower is (a) 25 m (b) 50√3 m (c) 25√m (d) \(\frac{50}{\sqrt{3}}\)m Answer: (c) 25√m

Question 10. From the top of a tower h metre high, the angle of depression of two objects, which lie on either side of it are a and p. The distance between the two objects is (a) h (cot α + cot β) (b) h (cot α – cot β) (c) h (tan α + tan β) (d) h (tan α – tan β) Answer: (a) h (cot α + cot β)

Fill in the blanks

Question 1. Line joining the eye of an observer to the object viewed by observer is called line of ___________ . Answer: sight

Question 2. The angle formed by the line of sight with the horizontal when it is above the horizontal level is called ___________ . Answer: angle of elevation

Question 3. The angle formed by the line of sight with the horizontal when it is below the horizontal level is called ___________ . Answer: angle of depression

Question 4. Trigonometry was invented because its need arose in ___________ . Answer: Astronomy

Question 5. If the length of the shadow of a tree is √3 times its height, then the angle of elevation is ___________ . Answer: 30°

Question 6. If the angle of elevation of a tree from a certain distance on the ground is 45°, then the height of tree is equal to ___________ . Answer: its distance from observation point

Question 7. The angles of elevation of the top of a tower from two points distant a and b from the base on the same straight line with it are complementary. The height of the tower is ___________ . Answer: √ab units

Question 8. The angle of ___________ of sun at a given time from a point is called sun’s ___________ at that time. Answer: elevation, altitude

Question 9. If the sun’s altitude increases, then the length of shadow of a tower ___________ . Answer: decreases

Question 10. A pole is being broken by a storm at one of its point of trisection, then the top struck the ground at an angle of ___________ . Answer: 30°

Extra Questions for Class 10 Maths

Ncert solutions for class 10 maths, free resources.

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

NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios. And you also learn what is the line of sight, angle of elevation, and angle of depression etc.  CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 9 Some Applications of Trigonometry Solutions

Below we have given the answers to all the questions present in Some Applications of Trigonometry in our NCERT Solutions for Class 10 Maths chapter 9. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

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Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

• Last modified on: 1 year ago
• Reading Time: 3 Minutes

Case Study Questions:

Question 1:

Ananya is feeling so hungry and so thought to eat something. She looked into the fridge and found a bread pieces. She decided to make a sandwich. She cut the piece of bread diagonally and found it forms a right-angled triangle, with sides 4 cm, 4√3 cm and 8 cm.

On the basis of above information, answer the following questions.

(i) The value of ∠M is

D. None of these

(ii) The value of ∠K is

(iii) Find the value of tan M.

(iv) sec 2 M – 1 =

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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

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NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

• Exercise 9.1

How many exercises in Chapter 9 Some Applications of Trigonometry

What is line of sight, what is angle of elevation, what is angle of depression, contact form.