case study of trigonometry class 10th

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

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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:

Answer: (d) 6m

(ii) Measure of ∠A =

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

Answer: (d) 2√2

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

Answer: (c) 2

Question 2:

Answer: (a) 30°

(ii) The measure of  ∠C is

Answer: (c) 60°

(iii) The length of AC is 

Answer: (d)6√3m

(iv) cos2A =

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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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:

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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

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Case Study Questions for Class 10 Maths Chapter 9 Applications of Trigonometry

• Last modified on: 10 months ago
• Reading Time: 9 Minutes

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 as well. In that, a paragraph will be given, and then the MCQ questions or subjective questions based on it will be asked.

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

Case Study Questions:

Question 1:

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 two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

(i) The distance of the satellite from the top of Mullayanagiri is

(a) 1139.4 km

(b) 577.52 km

(c) 1937 km

(d) 1025.36 km

(ii) If a mile stone very far away from, makes 45 to the top of Mullanyangiri mountain. So, find the distance of this milestone form the mountain.

(a) 1118.327 km

(b) 566.976 km

(iii) The distance of the satellite from the ground is

(iv) The distance of the satellite from the top of Nanda Devi is

(v) What is the angle of elevation if a man is standing at a distance of 7816m from Nanda Devi?

Question 2:

A group of students of class X visited India Gate on an educational 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 m) in height.

(i) What is the angle of elevation if they are standing at a distance of 42 m away from the monument? (a) 30° (b) 45° (c) 60° (d) 0°

(ii) They want to see the tower at an angle of 60°. So, they want to know the distance where they should stand and hence find the distance. (a) 25.24 m (b) 20.12 m (c) 42 m (d) 24.24 m

(iii) If the altitude of the Sun is at 60°, 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

(iv) The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is (a) 30° (b) 45° (c) 60° (d) 90°

(v) 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

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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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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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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
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

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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Introduction To Trigonometry Class 10 Notes

Cbse class 10 maths trigonometry notes:- download pdf here, class 10 maths chapter 8 introduction to trigonometry notes.

The notes for trigonometry Class 10 Maths are provided here. In maths, trigonometry is one of the branches where we learn the relationships between angles and sides of a triangle. Trigonometry is derived from the Greek words  ‘tri’ (means three), ‘gon’ (means sides) and ‘metron’ (means measure). In this chapter, we will learn the basics of trigonometry. Get the complete concept of trigonometry which is covered in Class 10 Maths. Also, get the various trigonometric ratios for specific angles, the relationship between trigonometric functions, trigonometry tables, and various identities given here.

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

• Introduction to Trigonometry Short Notes
• Introduction to Trigonometry MCQ Practice Questions
• Introduction to Trigonometry MCQ Practice Solutions

Trigonometric Ratios

Opposite & adjacent sides in a right-angled triangle.

In the Δ A B C right-angled at B, BC is the side opposite to ∠ A , AC is the hypotenuse, and AB is the side adjacent to ∠ A .

For the right Δ A B C , right-angled at ∠ B , the trigonometric ratios of the ∠ A are as follows:

• sin A=opposite side/hypotenuse=BC/AC
• cos A=adjacent side/hypotenuse=AB/AC
• tan A=opposite side/adjacent side=BC/AB
• cosec A=hypotenuse/opposite side=AC/BC
• sec A=hypotenuse/adjacent side=AC/AB
• cot A=adjacent side/opposite side=AB/BC

Relation between Trigonometric Ratios

• cosec θ =1/sin θ
• sec θ = 1/cos θ
• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ=1/tan θ

Example: Suppose a right-angled triangle ABC, right-angled at B such that hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm. Also, ∠ACB = θ. Find the trigonometric ratios tan θ, sin θ and cos θ.

Solution: Given, in ∆ABC,

Hypotenuse, AC = 5cm

Base, BC = 3cm

Perpendicular, AB = 4cm

Then, by the trigonometric ratios, we have;

tan θ = Perpendicular/Base = 4/3

Sin θ = Perpendicular/Hypotenuse = AB/AC = ⅘

Cos θ = Base/Hypotenuse = BC/AC = ⅗

To know more about Trigonometric Ratios, visit here .

Visualization of Trigonometric Ratios Using a Unit Circle

Draw a circle of the unit radius with the origin as the centre. Consider a line segment OP joining a point P on the circle to the centre, which makes an angle θ with the x-axis. Draw a perpendicular from P to the x-axis to cut it at Q.

• sin θ=PQ/OP=PQ/1=PQ
• cos θ=OQ/OP=OQ/1=OQ
• tan θ=PQ/OQ=sin θ/cos θ
• cosec θ=OP/PQ=1/PQ
• sec θ=OP/OQ=1/OQ
• cot θ=OQ/PQ=cos θ/sin θ

Trigonometric Ratios of Specific Angles

The specific angles that are defined for trigonometric ratios are 0°, 30°, 45°, 60° and 90°.

Trigonometric Ratios of 45°

If one of the angles of a right-angled triangle is 45°, then another angle will also be equal to 45°.

Let us say ABC is a right-angled triangle at B, such that;

∠ A = ∠ C = 45°

Thus, BC = AB = a (say)

Using Pythagoras theorem, we have;

AC 2 = AB 2 + BC 2

= a 2 + a 2

Now, from the trigonometric ratios, we have;

• sin 45° = (Opp. side to angle 45°)/Hypotenuse = BC/AC = a/a√2 = 1/√2
• cos 45° = (Adj. side to angle 45°)/Hypotenuse = AB/AC = a/a√2 = 1/√2
• tan 45° = BC/AB = a/a = 1
• cosec 45° = 1/sin 45° = √2
• sec 45° = 1/cos 45° = √2
• cot 45° = 1/tan 45° = 1

Trigonometric Ratios of 30° and 60°

Here, we will consider an equilateral triangle ABC, such that;

AB = BC = AC = 2a

∠A = ∠B = ∠C = 60°

Now, draw a perpendicular AD from vertex A that meets BC at D

According to the congruency of the triangle, we can say;

Δ ABD ≅ Δ ACD

∠ BAD = ∠ CAD (By CPCT)

Now, in triangle ABD, ∠ BAD = 30° and ∠ ABD = 60°

Using Pythagoras theorem,

AD 2 = AB 2 – BD 2

= (2a) 2 – (a) 2

So, the trigonometric ratios for a 30-degree angle will be;

sin 30° = BD/AB = a/2a = 1/2

cos 30° = AD/AB = a√3/2a = √3/2

tan 30° = BD/AD = a/a√3 = 1/√3

cosec 30° = 1/sin 30 = 2

sec 30° = 1/cos 30 = 2/√3

cot 30° = 1/tan 30 = √3

Similarly, we can derive the values of trigonometric ratios for 60°.

• sin 60° = √3/2
• cos 60° = 1/2
• tan 60° = √3
• cosec 60° = 2/√3
• sec 60° = 2
• cot 60° = 1/√3

Trigonometric Ratios of 0° and 90°

If ABC is a right-angled triangle at B, if ∠A is reduced, then side AC will come near to side AB. So, if ∠ A is nearing 0 degree, then AC becomes almost equal to AB, and BC get almost equal to 0.

Hence, Sin A = BC /AC = 0

and cos A = AB/AC = 1

tan A = sin A/cos A = 0/1 = 0

cosec A = 1/sin A = 1/0 = not defined

sec A = 1/cos A = 1/1 = 1

cot A = 1/tan A = 1/0 = not defined

In the same way, we can find the values of trigonometric ratios for a 90-degree angle. Here, angle C is reduced to 0, and the side AB will be nearing side BC such that angle A is almost 90 degrees and AB is almost 0.

Range of Trigonometric Ratios from 0 to 90 Degrees

For 0∘≤θ≤90∘,

• 0 ≤ sin θ ≤ 1
• 0 ≤ cos θ ≤ 1
• 0 ≤ tan θ < ∞ 1 ≤ sec θ < ∞
• 0 ≤ cot θ < ∞
• 1 ≤ cosec θ < ∞

tan θ and sec θ are not defined at  90∘.

cot θ and cosec θ are not defined at 0∘.

Variation of Trigonometric Ratios from 0 to 90 Degrees

As θ increases from 0 ∘ to 90 ∘

• s i n  θ increases from 0 to 1
• c o s  θ decreases from 1 to 0
• t a n  θ increases from 0 to ∞
• c o s e c  θ decreases from ∞ to 1
• s e c  θ increases from 1 to ∞
• c o t  θ decreases from ∞ to 0

Standard Values of Trigonometric Ratios

To know more about Trigonometric Ratios of Standard Angles, visit here .

Trigonometric Ratios of Complementary Angles

Complementary trigonometric ratios.

If θ is an acute angle, its complementary angle is 90 ∘ − θ . The following relations hold true for trigonometric ratios of complementary angles.

• s i n  ( 90° −  θ )  =  c o s  θ
• c o s  ( 90° −  θ )  =  s i n  θ
• t a n  ( 90° −  θ )  =  c o t  θ
• c o t  ( 90° −  θ )  =  t a n  θ
• c o s e c  ( 90° −  θ )  =  s e c  θ
• s e c  ( 90° −  θ )  =  c o s e c  θ

Example: Find the value of sin65°/cos25°.

Solution: Since,

cos A = sin (90° – A)

cos 25° = sin (90° – 25°)

Hence, sin65°/sin65° = 1

To know more about Trigonometric Ratios of Complementary Angles, visit here .

Trigonometric Identities

The three most important trigonometric identities are:

• s i n 2 θ + c o s 2 θ = 1
• 1 + c o t 2 θ = c o e s c 2 θ
• 1 + t a n 2 θ = s e c 2 θ

Example: Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Solution: We will start solving for LHS, to get RHS.

sec A (1 – sin A)(sec A + tan A) = (1/cos A)(1 – sin A)(1/cos A + sin A/cos A)

= [(1 – sin A)(1 + sin A)]/cos 2 A

= [1 – sin 2 A]/cos 2 A

= (cos 2 A)/(cos 2 A)

Hence proved.

To know more about Trigonometric Identities, visit here .

Related Articles

• NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
• Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs
• Important Questions for Class 10 Maths Chapter 8- Introduction to Trigonometry

Trigonometry for Class 10 Solved Problems

Find Sin A and Sec A, if 15 cot A = 8.

Given that 15 cot A = 8

Therefore, cot A = 8/15.

We know that tan A = 1/ cot A

Hence, tan A = 1/(8/15) = 15/8.

Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8

Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC, respectively.

Hence, BC = 15x and AB = 8x.

Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.

(i.e) AC 2 = AB 2 + BC 2

AC 2 = (8x) 2 +(15x) 2

AC 2 = 64x 2 +225x 2

AC 2 = 289x 2

Therefore, the hypotenuse AC = 17x.

Finding Sin A:

We know Sin A = Side Opposite to ∠A / Hypotenuse

Sin A = 15x/17x

Sin A = 15/17.

Finding Sec A:

To find Sec A, find cos A first.

Thus, cos A = Side adjacent to ∠A / Hypotenuse

Cos A = 8x/17x

We know that sec A = 1/cos A.

So, Sec A = 1/(8x/17x)

Sec A = 17x/8x

Sec A = 17/8.

Therefore, Sin A = 15/17 and sec A = 17/8.

If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that  0° <A+B ≤ 90°; A>B ]

Given that 

Tan (A+B) = √3.

We know that tan 60 = √3.

Thus, tan (A+B) = tan 60° = √3.

Hence A+B= 60° …(1)

Similarly, given that,

Tan (A-B) = 1/√3.

We know that tan 30° = 1/√3.

Thus, tan (A-B) = tan 30° = 1/√3.

Hence, A-B = 30° …(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60° + 30°

Now, substitute A = 45° in equation (1), we get

45° +B = 60°

B = 60°- 45°

Hence, A = 45 and B = 15°.

Video Lesson on Trigonometry

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Please visit: https://byjus.com/ncert-solutions-class-10-maths/chapter-8-introduction-to-trigonometry/

Very nice notes .It’s really help me

Can u answer my questions of this chapter!?

What’s the question

found these points becoming most helpful to solve my confusion very helpful notes thank you byju’s

Thank you!! This helped!

SinA=45° CosA=35° TanA=69° Please clear my doubt

cot full form please

Cot stands here for Cotangent

Cot stands for Cotangent

Really helpful thank you byju’s

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CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions 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) Value of DF is equal to

(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

(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) 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

(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?

(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

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

Cbse 10th standard maths subject some applications of trigonometry case study questions 2021 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) (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)

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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

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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CBSE Class 10th MCQs with Answers For Mathematics

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CBSE Class 10th MCQs with Answers For Mathematics PDF

It is very important to download CBSE Class 10th MCQs with Answers For Mathematics. It is so, because recently the CBSE board has introduced the 50% syllabus for MCQs Questions only. Therefore, in order to achieve good marks in the 10th Class Mathematics subject students are needed to practice the Mathematics MCQ Questions on a regular basis. For that reason we have provided the Mathematics MCQ Questions for Class 10 in PDF. 

Students can download them from the given link on this website.

Class 10 MCQs With Answers For Mathematics Chapter Wise Saves The Time

Class 10 MCQs With Answers For Mathematics Chapter Wise Saves The Time of students. Mathematics has quite complex topics and if any student wants to prepare the MCQs on their own, then they will likely waste a lot of their precious time.

But the MCQs for Mathematics that we have given here is completely based on the new syllabus. It is prepared by the subject matter experts of Mathematics. Also, the answers are provided to save more time. 

MCQ Questions with Answers For Improving the Speed and Accuracy

MCQ Questions also known as Multiple Choice Questions are a little confusing. Because one question contains four alternative options to choose the right one. At the time of answering those MCQ questions students can become puzzled to answer any one correct answer. 

Therefore the Mathematics MCQ will help them in answering such multiple choice questions. It will also aid in improving the accuracy of students. The answer will enable the students to quickly understand what will be the correct answer. 

However directly solving the MCQ questions with the help of Answer key will not be fruitful. It will be similar to mugging up things. Therefore, students are advised to use the MCQs Questions and Its answers to improve the accuracy and speed. 

Fastest Way to Answer the Mathematics MCQs Questions

While being under pressure and stress in the exam hall. Thinking about what would be the correct answer, etc. If any student will be aware of the fastest way to answer Mathematics MCQs Questions then their work will be very easy. Therefore the below given steps will help the students to know what is the fastest way to answer the Mathematics MCQs Questions.

• Read the Entire Question Carefully  : The biggest misconception in students is that they know the right answer before reading the entire question. To answer in the fastest way students are needed to read the questions very carefully. If required, then reading the questions multiple times will be quite helpful to answer the correct answer in the fastest way.
• Answer Them In Your Mind First  : Just after reading the questions, answer them in your mind first. It will save you from being confused in choosing the right answer.
• Use the Elimination Methods  : The elimination method is best to answer the Mathematics MCQ Questions. This method allows the students to reach the answer very quickly by eliminating all the wrong answers. 
• Focus on the Best Answer  : While answering the Mathematics MCQs, students start thinking about the answers that seem to be the correct ones. But in reality that is not necessarily the correct answer. Therefore, only focus on finding the best answers.
• Never Stick To Your First Choice   : Sticking to the first choice is sometimes a good choice. But not necessarily it is the only way to answer the questions in the fastest way. Student first choices may be wrong too. Therefore, never stick to your first choice. Always rely on your knowledge instead of your instinct.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
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