trigonometry case study questions class 10 chapter 9

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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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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.

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Case Study Questions Class 10 Maths Some Applications of Trigonometry

Case study questions class 10 maths chapter 9 some applications of trigonometry.

CBSE Class 10 Case Study Questions Maths Some Applications of Trigonometry. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Some Applications of Trigonometry.

CBSE Case Study Questions Class 10 Maths Some Applications of Trigonometry

Case Study – 1

[KVS Raipur 2021 – 22]

Answer – 20√3m

Answer – a) corresponding angle 

[ CBSE Academic Question Paper ]

(a) 1139.4 km

(d) 1025.36 km

Answer: (c) 1937 km

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
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Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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

August 7, 2019 by Sastry CBSE

Some Applications of Trigonometry Class 10 Important Questions Very Short Answer (1 Mark)

Some Applications of Trigonometry Class 10 Important Questions Short Answer-I (2 Marks)

Some Applications of Trigonometry Class 10 Important Questions Long Answer (4 Marks)

Question 33. The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. (2013OD) Solution: Refer to Question 30, Page 143.

Important Questions for Class 10 Maths

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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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CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions With Solution 2021

By QB365 on 22 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

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

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

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

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

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

(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

(v) Length of AC is equal to

(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.

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

Cbse 10th standard maths subject some applications of trigonometry case study questions with solution 2021 answer keys.

(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) (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):  \(\text { In } \Delta A D C\) \(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

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• 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
• Chapter 15 Probability

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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

• The solutions are designed by subject experts.
• Access to chapter-wise questions and answers.
• Valuable guidance for the students to prepare for the CBSE exams.
• The solutions are explained in detailed procedure.
• Easy access to all the exercise solutions.

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

Disclaimer – 

Dropped Topics –  9.1 Introduction

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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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Extra Questions of Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF Download

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We have provided you with Extra and Important Questions from Class 10 Maths Chapter 9 Some Applications of Trigonometry. This Extra and Important Questions will help you to score 100% in your Board Exams. These extra questions will be helpful to revise the important topics and concepts.

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Table of Contents

Some Applications of Trigonometry Class 10 Important Questions with Answers Maths Chapter 9

Extra questions for class 10 maths chapter 9 very short answer type.

If a man standing on a platform, 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Solution: False, θ 1  ≠ θ 1  (Fig. 11.15)

Question: Find the angle of elevation of the sun when the shadow of a pole h m high is √3 h m long.

Question: The height of a tower is 12 m. What is the length of its shadow when 10 Sun’s altitude is 45°?

Let AB be the tower [Fig. 11.17]. Then, ∠C = 45°, AB = 12 m

Question: 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° [Fig. 11.18].

Solution: Let AB be the vertical pole and AC be the long rope tied to point C. In right ∆ABC, we have

Therefore, height of the pole is 10 m.

Question: 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 BC be the tower whose height is h metres and A be the point at a distance of 30 m from the foot of the tower. The angle of elevation of the top of the tower from point A is given to be 30°. Now, in right angle ∆CBA we have,

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

Question: 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 OA be the tower of height h metre and P, l be the two points at distance of 9 m and 4 m respectively from the base of the tower. Now, we have OP = 9 m, OQ = 4 m, Let ∠APO = θ, ∠AQO = (90° – θ) and OA = h metre (Fig. 11.21) Now, in ∆POA, we have

Height cannot be negative. Hence, the height of the tower is 6 metre.

Extra Questions for Class 10 Maths Chapter 9 Short Answer Type

Question: Determine the height of a mountain if the elevation of its top at an unknown distance from the base is 30° and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 15o. (Use tan 15° = 0.27)

Let AB be the mountain of height h kilometres. Let C be a point at a distance of x km, from the base of the mountain such that the angle of elevation of the top at C is 30°. Let D be a point at a distance of 10 km from C such that angle of elevation at D is of 15°. In MBC (Fig. 11.22), we have

Substituting x = √3h in equation (i), we get ⇒ 0.27 ( √3h + 10) = h = 0.27 × √3h + 0.27 × 10 = h ⇒ 2.7 = h – 0.27 × √3h ⇒ 27 = h (1 – 0.27 × √3) ⇒ 27 = h (1 – 0.46) ⇒ h = 2.7/0.54 = 5 Hence, the height of the mountain is 5 km.

Question: The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

In Fig. 11.23, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°. Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC. So, BD = (40 + x) m Now, we have two right triangles ABC and ABD.

Using (i) in (ii), we get (x √3 ) √3 = x + 40, i.e., 3x = x + 40 i.e., x = 20 So, h = 20 √3 [From (i)] Therefore, the height of the tower is 20 √3 m.

Question: From a point P on the ground, the angle of elevation of the top of a 10m tall building is 30°. A flag is hosted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45o. Find the length of the flagstaff and the distance of the building from the point P. (You may take √3 = 1.732).

In Fig. 11.24, AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., BD and the distance of the building from the point P, i.e., PA.Since, we know the height of the building AB, we will first consider the right ∆PAB.

i.e., x = 100(√3 – 1) = 7.32 So, the length of the flagstaff is 7.32 m.

Question: 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.

In right angle ∆ABC, AC is the broken part of the tree (Fig. 11.20). So, the total height of tree = (AB + AC) Now in right angle ∆ABC,

Question: 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.

Let AB be the horizontal ground and K be the position of the kite and its height from the ground is 60 m and let length of string AK be x m. (Fig. 11.26) ∠KAB = 60° Now, in right angle ∆ABK we have

So, the length of string is 40 √3 m.

Question : 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 AB be the building and PQ be the initial position of the boy (Fig. 11.27) such that ∠APR = 30° and AB = 30 m Now, let the new position of the boy be P’Q’ at a distance QQ’. Here, ∠AP’R = 60° Now, in ∆ARP, we have

Therefore, required distance, QQ = PP’ = PR – P’R = 28.5 √3 – 9.5 √3 = 19√3 Hence, distance walked by the boy is 19√3 m.

Question: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

In Fig. 11.28, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3m. We are interested to determine the width of the river, which is the length of the side AB of the ∆APB. In right ∆ADP, ∠A = 30°

∴ DB = 3m Now, AB = BD + AD = 3 + 3 √3 = 3 (1 + √3) m Therefore, the width of the river is 3(√3 + 1) m.

Question: Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. (Use √3 = 1.73]

Let the distance between the two ships be d. Let the distance of one ship from the light house be x metres. Then, the distance of the other ship from the light house will be (d – x) metres.

Question: The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000 √3 m, find the speed of the aeroplane.

Let A be point of observation and P and Q be positions of the plane. Let ABC be the line through A and it is given that angles of elevation from point A to two positions P and Q are 60° and 30°. ∠PAB = 60°, ∠QAB = 30° Height = 3000 √3 m So, in ∆ABP, we have

Extra Questions for Class 10 Maths Chapter 9 Long Answer Type

Question: 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 AB be a building of height 20 m and BC be the transmission tower of height x m and D be any point on the ground (Fig. 11.36). Here, ∠BDA = 45° and ∠ADC = 60° Now, in ∆ADC, we have

⇒ x = 20√3 – 20 = 20 (√3 – 1) = 20 (1.732 – 1) = 20 × 0.732 = 14.64 m Hence, the height of tower is 14.64 m.

Question: 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 bottom of the pedestal is 45°. Find the height of the pedestal.

Let AB be the pedestal of height h metres and BC be the statue of height 1.6 m (Fig. 11.37). Let D be any point on the ground such that, ∠BDA = 45° and ∠CDA = 60° Now, in ∆BDA, we have

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

Question: 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° (Fig. 11.40). Find the distance travelled by the balloon during the interval.

Let A and B be two positions of the balloon and G be the point of observation. (eyes of the girl) Now, we have AC = BD = BQ – DQ = 88.2 m – 1.2 m = 87 m . ∠AGC = 60°, ∠BGD = 30° Now, in ∆AGC, we have

Hence, the balloon travels 58 √3 metres.

Question: 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 OA be the tower of height h, and P be the initial position of the car when the angle of depression is 30°. After 6 seconds, the car reaches to such that the angle of depression at Q is 60°. Let the speed of the car be v metre per second. Then, PQ = 6υ (∵ Distance = speed × time) and let the car take t seconds to reach the tower OA from Q (Fig. 11.41). Then, OQ = υt metres. Now, in ∆AQO, we have

Hence, the car will reach the tower from Q in 3 seconds.

Question: In Fig. 11.42, ABDC is a trapezium in which AB || CD. Line segments RN and LM are drawn parallel to AB such that AJ = JK = KP. If AB = 0.5 m and AP = BQ = 1.8 m, find the lengths of AC, BD, RN and LM.

We have, AP = 1.8 m AJ = JK = KP = 0.6 m AK = 2AJ = 1.2 m In ∆ARJ and ∆BNJ’ we have AJ = BJ, ∠ARJ = ∠BNJ = 60° and ∠AJR = ∠BJ’N = 90° ∴ ∆ARJ ≅ ∆BNJ ⇒ RJ = NJ (By AAS congruence criterion) Similarly, ∆ALK ≅ ∆BMK” ⇒ LK = MK” In ∆ARJ,

Since ∆ACP ≅ ∆BDQ So, BD = AC = 2.0784 m Now, RN = RJ + JJ + J’N = 2RJ + AB [∵ RJ = J’ N and JJ = AB] = 2 × 0.3464 +0.5 = 1.1928 m Length of step LM = LK + KK + KM = 2LK + AB [∵ LK = K M and KK = AB] = 2 × 0.6928 + 0.5 = 1.8856 m Thus, length of each leg = 2.0784 m = 2.1 m Length of step RN = 1.1928 m = 1.2 m and, length of step LM = 1.8856 m = 1.9 m

Question: Two poles of equal heights are standing opposite to 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 two poles of equal height h metre and let P be any point between the poles, such that ∠APB = 60° and ∠DPC = 30° The distance between two poles is 80m.(Given) Let AP = x m, then PC = (80 – x) m. h’m Now, in ∆APB, we have

Now, putting the value of x in equation (i), we have h = √3 × 20 = 20 √3 Hence, the height of the pole is 20 √3 m and the distance of the point from first pole is 20 m and that of the second pole is 60 m.

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

• Exercise 9.1

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Chapter 8 Class 10 Introduction to Trignometry

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The chapter is updated according to the new NCERT, for 2023-2024 Board Exams.

Get NCERT Solutions with videos of all questions and examples of Chapter 8 Class 10 Trigonometry. Videos of all questions are made with step-by-step explanations. Check it out now.

Trigonometry means studying relationship between measures of triangle. Usually, we talk about right triangles when we study trigonometry,

In this chapter, we will study

• What is sin, cos, tan ( Sine, cosine, tangent) ... and how they are found in a triangle
• What is sec, cosec, cot, and how is it related to sin, cos, tan.
• (Sin, cos, tan, sec, cosec, cot are known as Trigonometric Ratios)
• Then, we study Trigonometric ratios of specific angles l ike 0°, 30°, 45°, 60°, 90° ; and do some questions
• We study the formulas of sin (90 - θ) , cos (90 - θ), tan (90 - θ)
• And then we study Trigonometric Identities, and how other identities are derived from sin 2 θ + cos 2 θ = 1

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