case study questions class 10 introduction to trigonometry

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:

case study questions class 10 introduction to trigonometry

(a) 2m

(b) 3m

(d) 6m

Answer: (d) 6m

(ii) Measure of ∠A =

(a) 30°

(b) 60°

(d) None of these

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

(a) 0

(b) 1

(d) 2√2

Answer: (d) 2√2

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

(a) 0

(b) 1

(d) 1/2

Answer: (c) 2

Question 2:

(a) 30°

(b) 45°

(d) None of these

Answer: (a) 30°

(ii) The measure of ∠C is

Answer: (c) 60°

(iii) The length of AC is

(a)2√3 m

(b)√3m

(d)6√3m

Answer: (d)6√3m

(iv) cos2A =

(a) 0

(b)1/2

(d)√3/2

Answer: (b)1/2

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

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Case Study on Introduction to Trigonometry Class 10 Maths PDF

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

The case study questions on Introduction to Trigonometry are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Introduction to 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 Introduction to 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 Introduction to Trigonometry, therefore, they prepared a set of solutions along with the case study questions on Introduction to Trigonometry.

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

Why Solve Introduction to Trigonometry Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Introduction to 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 Introduction to Trigonometry Case study questions as it will help better prepare for the Class 10 board exam preparation.
Develop Problem-Solving Skills: Class 10 Maths Introduction to 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 Introduction to Trigonometry Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Introduction to Trigonometry as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Introduction to Trigonometry?

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

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

Basic Formulas of Introduction to Trigonometry: One of the most important things to know to solve Case Study Questions on Class 10 Introduction to Trigonometry is to learn about the basic formulas or revise them before solving the case-based questions on Introduction to 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 Introduction to Trigonometry case study questions.
Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Introduction to Trigonometry Class 10 Maths?

Use Selfstudys.com to find Case Study on Introduction to Trigonometry Class 10 Maths. For ease, here is a step-wise procedure to download the Introduction to Trigonometry Case Study for Class 10 Maths in PDF for free of cost.

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

Last modified on: 11 months ago
Reading Time: 3 Minutes

Case Study Questions:

Question 1:

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

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

(i) The value of ∠M is

D. None of these

(ii) The value of ∠K is

(iii) Find the value of tan M.

(iv) sec 2 M – 1 =

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CBSE Class 10 Maths: Case Study Questions of Chapter 8 Introduction to Trigonometry PDF Download

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 case study-based questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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

Introduction to Trigonometry Case Study Questions With answers

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

Case Study/Passage-Based Questions

Question 1:

(a) 2m

(b) 3m

(d) 6m

Answer: (d) 6m

(ii) Measure of ∠A =

(a) 30°

(b) 60°

(d) None of these

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

(a) 0

(b) 1

(d) 2√2

Answer: (d) 2√2

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

(a) 0

(b) 1

(d) 1/2

Answer: (c) 2

Question 2:

(a) 30°

(b) 45°

(d) None of these

Answer: (a) 30°

(ii) The measure of ∠C is

Answer: (c) 60°

(iii) The length of AC is

(a)2√3 m

(b)√3m

(d)6√3m

Answer: (d)6√3m

(iv) cos2A =

(a) 0

(b)1/2

(d)√3/2

Answer: (b)1/2

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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 Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

Real Numbers Case Study Question
Polynomials Case Study Question
Pair of Linear Equations in Two Variables Case Study Question
Quadratic Equations Case Study Question
Arithmetic Progressions Case Study Question
Triangles Case Study Question
Coordinate Geometry Case Study Question
Introduction to Trigonometry Case Study Question
Some Applications of Trigonometry Case Study Question
Circles Case Study Question
Area Related to Circles Case Study Question
Surface Areas and Volumes Case Study Question
Statistics Case Study Question
Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.


I	NUMBER SYSTEMS	06
II	ALGEBRA	20
III	COORDINATE GEOMETRY	06
IV	GEOMETRY	15
V	TRIGONOMETRY	12
V	MENSURATION	10
VI	STATISTICS & PROBABILITY	11

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

Find the production in the 1 st year.
Find the production in the 12 th year.
Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

Find the distance between Lucknow (L) to Bhuj(B).
If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

Find the distance PA.
Find the distance PB
Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

What is the length of the line segment joining points B and F?
The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

If the first circular row has 30 seats, how many seats will be there in the 10th row?
For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

$case study questions class 10 introduction to trigonometry$

Draw a neat labelled figure to show the above situation diagrammatically.

$case study questions class 10 introduction to trigonometry$

What is the speed of the plane in km/hr.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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10th Class Mathematics Introduction to Trigonometry Question Bank

Done case based (mcqs) - introduction to trigonometry total questions - 35.

Direction: Q. 1 to 5

Anika is studying in X standard. She is making a figure to understand trigonometric ratio shown as below.

In \[\Delta PQR,\] \[\angle Q\] is right angled, \[\Delta QTR\]is right angled at T and \[\Delta QST\] is right angled at S. \[PQ=12cm,\] \[QR=8.5cm,\] \[\text{ST}=\text{4 cm},\] \[\text{SQ}=5\text{ cm},\] \[\angle QTS=x{}^\circ \]and\[\angle TPQ=y{}^\circ \]. Give answers to her questions by looking at the figure.

Based on the above information give the answer of the following questions:

A) \[8\, cm\] done clear

B) \[\sqrt{65}\,cm\] done clear

C) \[7.5\,cm\] done clear

D) \[\sqrt{69}\,cm\] done clear

question_answer 2) The value of \[\tan x{}^\circ \] is:

A) \[\frac{7.5}{13}\] done clear

B) \[\frac{5}{4}\] done clear

C) \[\frac{4}{5}\] done clear

D) \[\frac{13}{7.5}\] done clear

question_answer 3) The value of \[\sec x{}^\circ \] is:

A) \[\frac{\sqrt{91}}{6}\] done clear

B) \[\frac{\sqrt{71}}{6}\] done clear

C) \[\frac{\sqrt{41}}{4}\] done clear

D) \[\frac{\sqrt{31}}{5}\] done clear

question_answer 4) The value of \[\sin y{}^\circ \] is:

A) \[\frac{4}{\sqrt{65}}\] done clear

B) \[\frac{4}{7}\] done clear

C) \[\frac{7}{4}\] done clear

D) \[\frac{\sqrt{65}}{7}\] done clear

question_answer 5) The value of \[\cot y{}^\circ \] is:

A) \[\frac{7}{4}\] done clear

C) \[\frac{\sqrt{65}}{4}\] done clear

Direction: Q. 6 to 10

A sailing boat with triangular masts is shown below. Two right triangles can be observed. Triangles PQR and PQS. both right-angled at Q. The distance \[\text{QR}=\text{2 m}\] and \[\text{QS}=\text{3 m}\] and height \[\text{PQ}=\text{5 m}\].

Based on the above information, give the answer of the following questions:

A) \[\frac{\sqrt{34}}{5}\] done clear

B) \[\frac{\sqrt{34}}{3}\] done clear

C) \[\frac{5}{3}\] done clear

D) \[\frac{3}{\sqrt{34}}\] done clear

question_answer 7) The value of cosec R is:

A) \[\frac{\sqrt{29}}{5}\] done clear

B) \[\frac{\sqrt{29}}{2}\] done clear

C) \[\frac{2}{5}\] done clear

D) \[\frac{5}{\sqrt{29}}\] done clear

question_answer 8) The value of \[\tan S+\cot R\] is:

A) \[\frac{9}{4}\] done clear

B) \[\frac{5}{3}\] done clear

D) \[\frac{31}{15}\] done clear

question_answer 9) Value of \[{{\sin }^{2}}R-{{\cos }^{2}}S\] is:

A) \[0\] done clear

B) \[1\] done clear

C) \[\frac{97}{85}\] done clear

D) \[\frac{589}{986}\] done clear

question_answer 10) Value of \[{{\sin }^{2}}S-{{\cos }^{2}}R\] is:

Direction: Q. 11 to 15

Three friends - Sanjeev, Amit and Digvijay are playing hide and seek in a park. Sanjeev and Amit hide in the shrubs and Digvijay have to find both of them. If the positions of three friends are at A, B and C respectively as shown in the figure and forms a right angled triangle such that \[\text{AB}=\text{9 m},\] \[\text{BC}=3\sqrt{3}\text{ m}\] and \[\angle B=90{}^\circ \].

Based on the above information, give the answer of the following questions:

A) \[30{}^\circ \] done clear

B) \[45{}^\circ \] done clear

C) \[60{}^\circ \] done clear

D) None of these done clear

question_answer 12) The measure of \[\angle C\] is:

question_answer 13) The length of AC is:

A) \[2\sqrt{3}\,m\] done clear

B) \[\sqrt{3}\,m\] done clear

C) \[4\sqrt{3}\,m\] done clear

D) \[6\sqrt{3}\,m\] done clear

question_answer 14) \[\cos 2A=\]

B) \[\frac{1}{2}\] done clear

C) \[\frac{1}{\sqrt{2}}\] done clear

D) \[\frac{\sqrt{3}}{2}\] done clear

question_answer 15) \[\sin \left( \frac{C}{2} \right)=\]

Direction: Q. 16 to 20

Priyanshi's daughter is feeling hungry so she thought of eating something. She looked into the fridge and found some bread pieces. She decided to make a sandwich. She cut the piece of bread diagonally and found that it forms a right angled triangle, with sides \[\text{4 cm},\] \[\text{4}\sqrt{3}\text{ cm}\]and\[\text{8 cm}\].

Based on the above information, give the answer of the following questions:

B) \[60{}^\circ \] done clear

C) \[45{}^\circ \] done clear

question_answer 17) The value of \[\angle K=\]

A) \[45{}^\circ \] done clear

B) \[30{}^\circ \] done clear

question_answer 18) Find the value of sec M.

A) \[\sqrt{3}\] done clear

B) \[\frac{2}{\sqrt{3}}\] done clear

C) \[1\] done clear

question_answer 19) \[{{\sec }^{2}}M-1=\]

A) \[\tan M\] done clear

B) \[\tan 2M\] done clear

C) \[{{\tan }^{2}}M\] done clear

question_answer 20) The value of \[\frac{{{\tan }^{2}}45{}^\circ -1}{{{\tan }^{2}}45{}^\circ +1}\]is:

C) \[2\] done clear

D) \[-1\] done clear

Direction: Q. 21 to 25

Soniya and her father went to meet her friend Ruhi for a party. When they reached Ruhi's place, Soniya saw the roof of the house, which is triangular in shape. She imagined the dimensions of the roof as given in the figure.

Based on the above information, give the answer of the following questions:

A) \[2m\] done clear

B) \[3m\] done clear

C) \[4m\] done clear

D) \[6m\] done clear

question_answer 22) Measure of \[\angle A=\]

question_answer 23) Measure of \[\angle C=\]

question_answer 24) Find the value of \[\sin A+\cos C.\]

D) \[\sqrt{2}\] done clear

question_answer 25) Find the value of \[{{\tan }^{2}}C+{{\tan }^{2}}A\].

D) \[\frac{1}{2}\] done clear

Direction: Q. 26 to 30

In structural design a structure is composed of triangles that are interconnecting. A truss is one of the major types of enginnering structures and is especially used in the design of bridges and buildings. Trusses are designed to support loads, such as the weight of people. A truss is exclusively made of long, straight members connected by joints at the end of each member.

This is a single repeating triangle in a truss system.

Based on the above information give the, answer of the following questions:

A) \[5\,ft\] done clear

B) \[6\,ft\] done clear

C) \[8\,ft\] done clear

D) \[\frac{8}{\sqrt{3}}\,ft\] done clear

question_answer 27) In above triangle, what is the length of BC?

D) \[4\sqrt{3}\,ft\] done clear

question_answer 28) If \[\text{sin A}=\text{sin C},\]what will be the length of BC?

A) \[2ft\] done clear

B) \[4ft\] done clear

C) \[8ft\] done clear

D) \[4\sqrt{2}ft\] done clear

question_answer 29) If \[\sin (A+B)=\frac{1}{\sqrt{2}}\]and \[\cos (A-B)=\frac{\sqrt{3}}{2},\] \[0{}^\circ <A+B\le 90{}^\circ ,\] \[A>B\]then \[\angle A\]is:

B) \[37.5{}^\circ \] done clear

C) \[32.5{}^\circ \] done clear

D) \[35{}^\circ \] done clear

question_answer 30) If the length of AB doubles what will happen to the Length of AC?

A) remains same done clear

B) doubles the original lengthy done clear

C) become three times the original length done clear

D) become half of the original length done clear

Direction: Q. 31 to 35

An electrician wanted to repair a street lamp at a height of 15 feet. He places his ladder such that its foot is 8 feet from the foot of the lamp post as shown in the figure below:

Based on the above information give the, answer of the following questions:

A) \[\frac{8}{15}\] done clear

B) \[\frac{8}{17}\] done clear

C) \[\frac{15}{8}\] done clear

D) \[\frac{15}{17}\] done clear

question_answer 32) The value of \[\text{cosec}\,\,\text{P}\] is:

A) \[\frac{8}{17}\] done clear

B) \[\frac{15}{17}\] done clear

C) \[\frac{17}{8}\] done clear

D) \[\frac{17}{15}\] done clear

question_answer 33) The value of \[\frac{\sin R-\cos P}{\sin R+\cos P}\] is:

A) \[\frac{17}{30}\] done clear

B) \[\frac{30}{17}\] done clear

C) \[0\] done clear

D) \[1\] done clear

question_answer 34) The value of cot P is:

A) \[\frac{15}{17}\] done clear

B) \[\frac{17}{15}\] done clear

C) \[\frac{8}{15}\] done clear

D) \[\frac{15}{8}\] done clear

question_answer 35) The value of \[\tan R+\frac{3}{\sec P}-1\]is:

A) \[\frac{253}{136}\] done clear

B) \[\frac{357}{136}\] done clear

C) \[\frac{479}{136}\] done clear

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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 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

(ii) The measure of $\angle$ C is

(iii) The length of AC is

$(c) 4 \sqrt{3} \mathrm{~m}$

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

(iv) cos2A =

(a) 0

$(b) \frac{1}{2}$

$(c) \frac{1}{\sqrt{2}}$

$(d) \frac{\sqrt{3}}{2}$

(v) sin $\left(\frac{C}{2}\right)$ =

$(a) \frac{1}{2}$

$(b) \frac{1}{\sqrt{2}}$

$(c) \frac{\sqrt{3}}{2}$

$(d) \frac{3}{4}$

(ii) Find cot B

$(a) \frac{3}{4}$

$(b) \frac{15}{4}$

$(c) \frac{3}{8}$

$(d) \frac{15}{8}$

(iii) Find tanA.

$(a) 2$

$(b) \sqrt{2}$

$(c) \frac{4}{3}$

$(d) \frac{2}{\sqrt{3}}$

(iv) Find secA.

$(a) 1$

$(b) \frac{2}{3}$

$(c) \frac{4}{3}$

$(d) \frac{5}{3}$

(v) Find cosecB.

$(a) \frac{17}{8}$

$(b) \frac{12}{5}$

$(c) \frac{5}{12}$

$(d) \frac{8}{17}$

$(a) \frac{12}{5}$

$(b) \frac{5}{12}$

$(c) \frac{12}{13}$

$(d) \frac{13}{12}$

(ii) The value of sec $\theta$ =

$(a) \frac{5}{12}$

$(b) \frac{12}{5}$

$(c) \frac{13}{12}$

$(d) \frac{12}{13}$

(iii) The value of $\frac{\tan \theta}{1+\tan ^{2} \theta}=$

$(a) \frac{5}{12}$

$(b) \frac{12}{5}$

$(c) \frac{60}{169}$

$(d) \frac{169}{60}$

(iv) The value of $\cot ^{2} \theta-\operatorname{cosec}^{2} \theta=$

(v) The value of $\sin ^{2} \theta+\cos ^{2} \theta=$

(ii) The value of $\angle$ K =

(iii) Find the value of tanM.

$(a) \sqrt{3}$

$(b) \frac{1}{\sqrt{3}}$

(iv) sec 2 M - 1 =

(v) The value of $\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}$ is

(ii) Measure of $\angle$ A =

(iii) Measure of $\angle$ C =

(iv) Find the value of sinA + cosC.

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

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

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Important Questions Class 10 Mathematics Chapter 8

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Important Questions Class 10 Mathematics Chapter 8 – Introduction to Trigonometry

Mathematics requires analytical thinking and problem-solving skills. One should do lots of practice and develop a deep understanding of the concepts of the subject in order to excel in Mathematics. As a result, students should have the right resources to refer to and follow accurate strategies. Students are advised to solve the Important Questions Class 10 Mathematics Chapter 8 as it caters to varying levels of difficulty from easy to challenging questions covering all the topics from the chapter.

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Introduction to Trigonometry is a chapter that covers the basics of Trigonometry. The chapter covers the ratio of the sides of a right triangle with respect to its acute angle and the trigonometric ratios for angles of measure 0° and 90°. Chapter 8 Class 10 Mathematics important questions will cover questions from the topics below:

Trigonometric Ratios
Trigonometric Ratios of Some Specific Angles
Trigonometric Ratios of Complementary Angles
Trigonometric Identities

The important questions of Class 10 Mathematics Chapter 8 are made by Mathematics faculty experts who understand the key topics which students might find challenging while studying, therefore these questions are prepared in such a way that they can get to the point answers without wasting much time on a single subject. As a result, students can get a clear understanding of all the topics and improve their performance by solving these questions.

Extramarks is a trustworthy and reliable source for all NCERT-related study material. Students can get the NCERT textbook, NCERT exemplar, NCERT solutions, revision notes, CBSE mock tests, etc. on our official website. In fact, these resources are the best study material for getting a 100% score and creating your first milestone in high school.

Important Questions Class 10 Mathematics Chapter 8 – With Solutions

The following important questions and their solutions are included in the Mathematics Class 10 Chapter 8 important questions:

Question 1. when cos A = 4/5, the value for tan A is

Answer 1. (B) 3/4

Explanation: According to the question,

cos A = 4/5 …(1)

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin 2 θ + cos 2 θ =1

So, sin θ = √ (1- cos 2 θ)

sin A = √ (1- cos 2 A) …(2)

sin 2 A = 1- cos 2 A

sin A = √(1- cos 2 A)

Substituting equation (1) in (2),

Sin A = √(1-(4/5) 2 )

= √(1-(16/25))

Tan A = (3/5)*(5/4) = (3/4)

Question 2. when sin A = ½, the value of cot A is

Answer 2. (A) √3

Sin A = ½ … (1)

We know that,

Cot A = 1/tan A = cos A/sin A …(2)

To find the value of cos A.

So, cos θ = √(1-sin 2 θ)

cos A = √(1-sin 2 A) … (3)

cos2 A = 1-sin 2 A

cos A = √ (1-sin 2 A)

Substituting equation 1 in 3, we get,

cos A = √(1-1/4) = √(3/4) = √3/2

here, Substituting values of sin A and cos A in the equation 2, we get

cot A = (√3/2) × 2 = √3

Question 3. The value of the expression are as [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

Answer 3. (B) 0

Explanation: As per question,

we have to find out the value of the equation as,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

hence, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We observe,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

Question 4. Here, sinθ = a b , so cosθ is equal to

(A) a/√(b2– a2)

(D) b/√(b2-a2)

Answer 4. (C) √( b 2 – a 2 )/b

We know, sin 2 θ +cos 2 θ =1

sin A = √(1- cos 2 A

thus, cos θ = √(1- a 2 / b 2 ) = √(( b 2 – a 2 )/ b 2 ) = √( b 2 – a 2 )/b

therefore, cos θ = √( b 2 – a 2 )/b

Question 5. If cos (α + β) = 0, then sin (α – β) can be reduced to

Answer 5. (B) cos 2β

cos(α+β) = 0

hence, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing the cosine equation on L.H.S and R.H.S,

Now we need to reduce sin (α -β ),

thus, we use,

sin(α-β) = sin(90°-β-β) = sin(90°-2β)

sin(90°-θ) = cos θ

thus, sin(90°-2β) = cos 2β

Therefore, sin(α-β) = cos 2β

Question 6. The value of the equation (tan1° tan2° tan 3 ° … tan89°) is

Answer 6. (B) 1

Explanation: tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

hence, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

Since, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

Since, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44°). (1/tan 43°)… (1/tan 3°). (1/tan 2°). (1/tan 1°)

=(tan1° * 1/tan1°).(tan 2° * 1/tan 2°)…(tan 44° * 1/tan 44°)

thus, tan 1°.tan 2°.tan 3° …… tan 89° = 1

Question 7. If cos 9α = sinα and 9α < 90°, then the value of tan5α is

Answer 7. (C) 1

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1

Question 8. tan 47°/cot 43° = 1

Answer 8.

Justification:

Since, tan (90° -θ) = cot θ

(tan 47°/tan 43°) = tan(90° – 47°)/cot 43°

(tan 47°/tan 43°) = (cot 43°/cot 43°) = 1

(tan 47°/cot 43°) = 1

Question 9. The value for the expression (cos 2 23° – sin 2 67°) will be positive.

Answer 9.

Since, (a 2 -b 2 ) = (a+b)(a-b)

cos 2 23° – sin 2 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

Question 10. The value for the expression (sin 80° – cos 80°) is negative.

Answer 10.

clarification:

sin θ increases if 0° ≤ θ ≤ 90°

cos θ decreases if 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value – decreasing value)

= a positive value.

Thus, (sin 80°-cos 80°) > 0.

Question 11. If cosA + cos2A = 1, then sin2A + sin4A = 1.

Answer 11.

According to the question,

cos A+ cos 2 A = 1

i.e., cos A = 1- cos 2 A

sin 2 θ+ cos 2 θ = 1

sin 2 θ = 1- cos 2 θ)

cos A = sin 2 A …(1)

Squaring L.H.S and R.H.S,

cos 2 A = sin4 A …(2)

To find sin2A+sin4 A=1

Adding equations (1) and (2),

sin2A + sin4 A= cos A + cos 2 A

Therefore, sin2A+ sin4 A = 1

Question 12. (tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec 2 θ.

Answer 12.

L.H.S = (tan θ+2) (2 tan θ+1)

= 2 tan 2 θ + tan θ + 4 tan θ + 2

= 2 tan 2 θ+5 tan θ+2

Since, sec 2 θ – tan 2 θ = 1, we get, tan 2 θ = sec 2 θ-1

= 2(sec 2 θ-1) +5 tan θ+2

= 2 sec 2 θ-2+5 tan θ+2

= 5 tan θ+ 2 sec 2 θ ≠R.H.S

∴, L.H.S ≠ R.H.S

Question 13. (√3+1) (3 – cot 30°) = tan 3 60° – 2 sin 60°

Answer 13.

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3) 2 – 1) √3 [∵ (√3+1)(√3-1) = ((√3) 2 – 1)]

Similarly solving R.H.S: tan 3 60° – 2 sin 60°

Since, tan 60o = √3 and sin 60o = √3/2,

(√3)3 – 2.(√3/2) = 3√3 – √3

Therefore, L.H.S = R.H.S

Hence, proved.

Question 14. If 1 + sin 2 θ = 3sinθ cosθ , then prove that tanθ = 1 or ½.

Answer 14.

Given: 1+sin 2 θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin 2 θ,

(1+sin 2 θ)/sin 2 θ = 3 sin θ cos θ/ sin 2 θ

⇒ (1/sin 2 θ) + 1 = 3cos θ/ sin θ

cosec 2 θ + 1 = 3 cot θ

cosec 2 θ – cot 2 θ = 1 ⇒ cosec 2 θ = cot 2 θ +1

⇒ cot 2 θ +1+1 = 3 cot θ

⇒ cot 2 θ +2 = 3 cot θ

⇒ cot 2 θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot 2 θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

tan θ = 1/cot θ

tan θ = 1, ½

Question 15. Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Answer 15.

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ) 2 = 1

⇒ sin 2 θ + 4 cos 2 θ + 4sin θcos θ = 1

Since, sin 2 θ = 1 – cos 2 θ and cos 2 θ = 1 – sin 2 θ

⇒ (1 – cos 2 θ) + 4(1 – sin 2 θ) + 4sin θcos θ = 1

⇒ 1 – cos 2 θ + 4 – 4 sin 2 θ + 4sin θcos θ = 1

⇒ – 4 sin 2 θ – cos 2 θ + 4sin θcos θ = – 4

⇒ 4 sin 2 θ + cos 2 θ – 4sin θcos θ = 4

a 2 + b 2 – 2ab = ( a – b) 2

So, we get,

(2sin θ – cos θ) 2 = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

Question 16. In ∆ ABC, the right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Answer 16.

In the given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm as well as BC = 7 cm

According to the Pythagoras Theorem,

In the right-angled triangle, the squares of the hypotenuse sides are equal to the addition of the squares for the other two sides.

on applying the Pythagoras theorem, we observe

AC 2 =A b 2 +BC 2

AC 2 = (24) 2 +72

AC 2 = (576+49)

AC 2 = 625cm 2

AC = √625 = 25

thus, AC = 25 cm

(i) To find Sin (A), Cos (A)

We get that sine (or) Sin function is equal to the ratio of length of the opposite sides to the hypotenuse sides. So it becomes

Sin (A) = Opposite side /Hypotenuse side = BC/AC = 7/25

Cosine or Cos function is same as the ratio of the length of the adjacent side to the hypotenuse side so it becomes,

Cos (A) = Adjacent side/Hypotenuse side = AB/AC = 24/25

(ii) To find out Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

Question 17. In Fig. find tan P – cot R

Answer 17.

In the given triangle QPR, the given triangle is right angled at Q, and the given measures are:

Since the given triangle is the right-angled triangle, to find the side QR, apply the Pythagorean theorem

According to the Pythagorean theorem,

In the right-angled triangle, the squares of the hypotenuse side is equal to the addition of the squares for the other two sides.

PR 2 = QR 2 + PQ 2

Substitute the values for PR and PQ

13 2 = QR 2 +12 2

169 = QR 2 +144

hence, QR 2 = 169−144

QR = √25 = 5

hence, the side QR = 5 cm

To find tan P – cot R:

According to trigonometric ratio, the tangent function is same as the ratio of the length of the opposite sides to the adjacent sides, the value for tan (P) becomes

tan (P) = Opposite sides /Adjacent sides = QR/PQ = 5/12

hence, cot function is the reciprocal of the tan function, then the ratio of the cot function becomes,

Cot (R) = Adjacent sides/Opposite sides = QR/PQ = 5/12

tan (P) – cot (R) = 5/12 – 5/12 = 0

hence, tan(P) – cot(R) = 0

Question 18. If sin A = 3/4, so Calculate cos A and tan A.

Answer 18.

Let us assume that a right-angled triangle ABC, right angled at B

Given: Sin A = 3/4

We get that, Sin function is equal to the ratio of length of the opposite side to the hypotenuse side.

hence, Sin A = Opposite side /Hypotenuse side = 3/4

Let’s BC be 3k, and AC will be 4k

where k is a positive real number.

According to Pythagoras theorem, the squares of the hypotenuse side is same as the sum of the squares for the other two sides of the right angle triangle, and we observe,

AC 2 =A b 2 + BC 2

Substitute the value of AC and BC

(4k) 2 =A b 2 + (3k) 2

16k 2 −9k 2 =A b 2

A b 2 =7k 2

hence, AB = √7k

then, we have to find the value of cos A and tan A

We get that,

Cos (A) = Adjacent side/Hypotenuse side

here, Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we observe

AB/AC = √7k/4k = √7/4

hence, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the value of the line BC and AB and cancel the constant k in both numerator and denominator, we observe,

BC/AB = 3k/√7k = 3/√7

hence, tan A = 3/√7

Question 19. Given 15 cot A = 8, find out sin A and sec A.

Answer 19.

Let us assume a right-angled triangle ABC, right angled at B

Given: 15 cot A = 8

So, Cot A = 8/15

We get that cot function is equal to the ratio of length of the adjacent side to the opposite side.

Hence, cot A = Adjacent sides /Opposite sides = AB/BC = 8/15

Let AB be 8k as well as BC will be 15k

Here, k is the positive real number.

According to the Pythagoras theorem, so the squares of the hypotenuse side is same as the addition of the squares for the other two sides of the right angle triangle, and we observe,

Substitute the value of AB and BC

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

AC 2 = 64k 2 + 225k 2

AC 2 = 289k 2

Hence, AC = 17k

Then, we have to find the value of sin A and sec A

Sin (A) = Opposite side /Hypotenuse

Here, Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we observe

Sin A = BC/AC = 15k/17k = 15/17

Hence, sin A = 15/17

Hence, secant or sec function is the reciprocal of the cost function which is the same as the ratio for the length of the hypotenuse side to adjacent side.

Sec (A) = Hypotenuse/Adjacent side

here, Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we observe,

AC/AB = 17k/8k = 17/8

hence, sec (A) = 17/8

Question 20. Given sec θ = 13/12 Calculate all the other trigonometric ratios

Answer 20.

We get that sec function is the reciprocal of the cost function which is same as the ratio of the length of hypotenuse sides to the adjacent sides

sec θ =13/12 = Hypotenuse side/Adjacent side = AC/AB

Let’s AC be 13k, and AB will be 12k

here, k is a positive real number.

According to Pythagoras theorem, the squares of the hypotenuse side is equal to the addition of the squares for the other two sides of the right angle triangle and we observe,

here, Substitute the value of AB and AC

(13k) 2 = (12k) 2 + BC 2

169k 2 = 144k 2 + BC 2

BC 2 = 169k 2 – 144k 2

BC 2 = 25k 2

hence, BC = 5k

then, substitute the corresponding values in all other trigonometric ratios

Sin θ = Opposite Side/Hypotenuse Side = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse Side = AB/AC = 12/13

tan θ = Opposite Sides/Adjacent Sides = BC/AB = 5/12

Cosec θ = Hypotenuse Side/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Sides /Opposite Sides = AB/BC = 12/5

Question 21. If ∠A and ∠B are the acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Answer 21.

Let us assume that the triangle ABC in which CD⊥AB

Given that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cost ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

then consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying the Pythagoras theorem in △CAD and △CBD we get,

CD 2 = BC 2 – BD 2 … (3)

CD 2 =AC 2 −AD 2 ….(4)

From the equations (3) and (4) we observe,

AC 2 −AD 2 = BC 2 −BD 2

then substitute the equations (1) and (2) in (3) and (4)

k 2 (BC 2 −BD 2 )=(BC 2 −BD 2 ) k 2 =1

Putting this value in equation, we obtain that

∠A=∠B (Angles opposite to the equal side are equal-isosceles triangle)

Question 22. In triangle ABC, right-angled at B, when tan A = 1/√3 find out the value :

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Answer 22.

Let’s ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let’s BC = 1k and AB = √3 k,

here k is the positive real number of the problem

By the Pythagoras theorem in ΔABC we get:

AC 2 =(√3 k) 2 +(k) 2

AC 2 =3k 2 +k 2

then find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

now, find the values of cos C and sin C

Sin C = AB/AC = √ 3/2

Cos C = BC/AC = 1/2

then, substitute the values in the given problems

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = ( √ 3/2 )(1/2) – (1/2) ( √ 3/2 ) = 0

Question 23. In ∆ QPR, right-angled at Q, PR + QR = 25 cm as well as PQ = 5 cm. Determine the value of sin P, cos P and tan P

Answer 23.

In the given triangle PQR, right angled at Q, the following measures are

PR + QR = 25 cm

then let us assume, QR = x

hence, According to the Pythagorean Theorem,

PR 2 = PQ 2 + QR 2

Substitute the value of PR as x

(25- x) 2 = 5 2 + x 2

25 2 + x 2 – 50x = 25 + x 2

625 + x 2 -50x -25 – x 2 = 0

-50x = -600

x= -600/-50

x = 12 = QR

then, find the value of PR

PR = 25- QR

Substitute the value of QR

then, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse Side = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse Side = PQ/PR = 5/13

(3) Tan p =Opposite Sides/Adjacent sides = QR/PQ = 12/5

Question 24. State whether the following will be true or false. clarify your answer.

(i) The value for tan A is always less than 1.

(ii) sec A = 12/5 thus some value of the angle A.

(iii)cos A is the abbreviation used to the cosecant of the angle A.

(iv) cot A is a product of cot A.

(v) sin θ = 4/3 for some of the angles θ.

Answer 24.

(i) The value of tan A will always be less than 1.

Answer: False

Proof: In ΔMNC the angle ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 that is greater than.

The triangle can be formed with the sides equal to 3, 4 and hypotenuse = 5 according to the Pythagoras theorem.

MC 2 =MN 2 +NC 2

5 2 =3 2 +4 2

(ii) sec A = 12/5 thus, some value of the angle A

Answer: True

clarification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, here k is a positive real number.

By the Pythagoras theorem we observe,

(12k) 2 =(5k) 2 +NC 2

NC 2 +25k 2 =144k 2

NC 2 =119k 2

Such a triangle is possible as it would follow the Pythagoras theorem.

(iii) cos A is an abbreviation used for the cosecant of the angle A.

clarification: Abbreviation used for the cosecant of angle M is cosec M. cos M is the abbreviation used for the cosine of angle M.

(iv) cot A is the product of the cot and A.

clarification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.

Answer : False

clarification: sin θ = Opposite/Hypotenuse

We know that in the right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always be less than 1 and it can never be 4/3 for any value of θ.

Question 25. Choose the correct option and clarify your answer:

(i) 2tan 30°/1+tan230° =

(A) sin 60° (B) sin 45° (C) tan 60° (D) cos 30°

(ii) 1-tan245°/1+tan245° =

(A) tan 90° (B) 0 (C) sin 45° (D) 1

(iii) sin 2A = 2 sin A will true if A =

(A) 0° (B) 30° (C) 45° (D) 60°

(iv) 2tan30°/1-tan230° =

(A) cos 60° (B) sin 45° (C) tan 60° (D) sin 30°

Answer 25.

(i) (A) is correct.

here, Substitute for tan 30° in a given equations

tan 30° = 1/√3

2tan 30°/1+tan 2 30° = 2(1/√3)/1+(1/√3) 2

= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

we obtained that the solution is equivalent to the trigonometric ratio as sin 60°

(ii) (D) is correct.

Substitute the of tan 45° in given equation

tan 45° = 1

1-tan 2 45°/1+tan 2 45° = (1-12)/(1+12)

The solution of the above equations is 0.

(iii) (A) is correct.

To find out the value for A, substitute the degree given in the option one by one

sin 2A = 2 sin A is true if A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

Apply in the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

then, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to the cos value, i.e., cos 0 =1

Therefore, ⇒ A = 0°

(iv) (C) is correct.

Substitute the of tan 30° in given equations

2 tan 30°/1-tan 2 30° = 2(1/√3)/1-(1/√3) 2

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The value for the given equation will be equivalent to tan 60°.

Question 26. when tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

Answer 26.

tan (A + B) = √3

Since √3 = tan 60°

then, substitute the degree values

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equations (i)

tan (A – B) = 1/√3

hence, 1/√3 = tan 30°

Now substitute the degree values

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

then add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

now, substitute the value for A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

Therefore A = 45° and B = 15°

Question 27. State whether the following will be true or false. clarify your answers.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ will increase.

(iii) The value of cos θ increases as θ will increase.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Answer 27. (i) False.

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we observe

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Hence the values obtained are not equal, the solution is false.

As per the values obtained as per the unit circle, the values of sin will:

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Therefore, the value of sin θ increases as θ increases. thus, the statement is true

(iii) False.

As per the values obtained as per the unit circle, the values of cos will:

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Therefore, the value of cos θ decreases as θ will increase. So, the statement given above is false.

sin θ = cos θ, if a right triangle has 2 angles of (π/4). Thus, the above statement is false.

hence cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

then substitute A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Thus, it is true

Question 28. Evaluate :

(i) sin 18°/cos 72°

(ii) tan 26°/cot 64°

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

Answer 28. (i) sin 18°/cos 72°

To simplify it, convert the sin function into the cos function

We get that, 18° is written as 90° – 18°, which is same as the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify the equation

= cos 72° /cos 72° = 1

To simplify it, convert the tan function into cot function

We get that, 26° is written as 90° – 26°, which is equal to the cot 64°.

= tan (90° – 26°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify it, convert the cos function into the sin function

We get that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify it, convert the cosec function into the sec function

We get that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0

Question 29. Shown in the equation:

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Answer 29. (i) tan 48° tan 23° tan 42° tan 67°

here, we Simplify the given problem by converting some of the tan functions into the cot functions

We get that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute to the value

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

here, Simplify the given problem by converting some of the cos functions to the sin functions

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

here, Substitute the value

= sin 52° sin 38° – sin 38° sin 52° = 0

Question 30. when tan 2A = cot (A – 18°), where 2A is the acute angle, find the value of A .

Answer 30. tan 2A = cot (A- 18°)

We get that tan 2A = cot (90° – 2A)

Substitute the above equation in given problem

⇒ cot (90° – 2A) = cot (A -18°)

then, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

hence, the value of A = 36°

Question 31. If tan A = cot B, proved that A + B = 90°.

Answer 31. tan A = cot B

We get that cot B = tan (90° – B)

To proved A + B = 90°, substitute the above equation in given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

thus Proved.

Question 32. when sec 4A = cosec (A – 20°), here 4A is an acute angle, findout the value of A.

Answer 32. sec 4A = cosec (A – 20°)

We get that sec 4A = cosec (90° – 4A)

To find out the value of A, thus, substitute the above equation in a given problems

cosec (90° – 4A) = cosec (A – 20°)

then, equate the angles

90° – 4A= A- 20°

A = 110°/ 5 = 22°

hence, the value of A = 22°

Question 33. when A, B and C are interior angles of the triangle ABC, then show that

sin (B+C/2) = cos A/2

Answer 33. We get that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find out the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

then, multiply both sides by the sin functions, we observe

⇒ sin (B+C)/2 = sin (90°-A/2)

hence, sin (90°-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

therefore, proved.

Question 34. Express sin 67° + cos 75° in the terms of the trigonometric ratios of angles in between 0° and 45°.

Answer 34. Given:

sin 67° + cos 75°

In term of sin as the cos function and cos as sin function, it can be written as

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

then , simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is expressed as cos 23° + sin 15°

Question 35. Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.

Answer 35. F To convert the given trigonometric ratios in terms of cost functions, use trigonometric formulas

cosec 2 A – cot 2 A = 1

cosec 2 A = 1 + cot 2 A

hence, cosec function is the inverse of sin function, it is written as

1/sin 2 A = 1 + cot 2 A

then, rearrange the terms, it becomes

sin 2 A = 1/(1+cot 2 A)

then, take square roots on both sides, we get

sin A = ±1/(√(1+cot 2 A)

The above equation define that the sin function in terms of cot function

then, to express sec function in terms of cot function, use this formula

sin 2 A = 1/ (1+cot 2 A)

then, represent the sin function as cos function

1 – cos2A = 1/ (1+cot 2 A)

Rearrange the terms,

cos2A = 1 – 1/(1+cot 2 A)

⇒cos2A = (1-1+cot 2 A)/(1+cot 2 A)

hence, sec function is the inverse of cos function,

⇒ 1/sec2A = cot 2 A/(1+cot 2 A)

Take the reciprocal and the square roots on both sides, we get

⇒ sec A = ±√ (1+cot 2 A)/cotA

then, to express the tan function in the terms of the cot function

tan A = sin A/cos A as we cot A = cos A/sin A

hence, cot function is the inverse of tan function, thus, it is rewrite are as

tan A = 1/cot A

Question 36. Write the other trigonometric ratios of ∠A in the terms of sec A.

Answer 36. Cos A function in the terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in the terms of sec A:

cos 2 A + sin 2 A = 1

here, Rearrange the terms

sin 2 A = 1 – cos 2 A

sin 2 A = 1 – (1/sec 2 A)

sin 2 A = (sec 2 A-1)/sec 2 A

sin A = ± √(sec 2 A-1)/sec A

cosec A function in the terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec 2 A-1)

then, tan A function in terms of sec A:

sec 2 A – tan2A = 1

Rearrange the terms

⇒ tan 2 A = sec 2 A – 1

tan A = √(sec 2 A – 1)

cot A function in the terms for sec A:

⇒ cot A = 1/tan A

cot A = ±1/√(sec 2 A – 1)

Question 37. Evaluate that:

(i) (sin263° + sin227°)/(cos217° + cos273°)

(ii) sin 25° cos 65° + cos 25° sin 65°

Answer 37.

(i) (sin 2 63° + sin 2 27°)/( cos 2 17° + cos 2 73°)

To simplify it, convert some of the sin functions into the cos functions and cos function into the sin function and it become as,

= [sin 2 (90°-27°) + sin 2 27°] / [ cos 2 (90°-73°) + cos 2 73°)]

= ( cos 2 27° + sin 2 27°)/(sin 2 27° + cos 2 73°)

= 1/1 =1 (hence sin 2 A + cos2A = 1)

hence, (sin 2 63° + sin 2 27°)/( cos 2 17° + cos 2 73°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

for simplify it, convert some of the sin functions into the cos functions and the cos function into sin function and it become as,

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos 2 65° + sin 2 65° = 1 (since sin 2 A + cos2A = 1)

hence, sin 25° cos 65° + cos 25° sin 65° = 1

Benefits of Solving Important Questions Class 10 Mathematics Chapter 8

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Prove that 1 + cosA 1 cosA = cosecA + cotA

Marks: 3 Ans

1 + cosA 1 cosA = 1 + cosA × 1 + cosA 1 cosA × 1 + cosA = 1 + cosA 2 1 cos 2 A = 1 + cosA 2 sin 2 A ( 1 = cos 2 A + sin 2 A ) = 1 + cosA sinA = cosecA + cotA

Q.2 If sinx = 1, then find other trigonometric ratios.

Marks: 2 Ans

Since, cos x = 1 sin 2 x = 1 1 2 = 0 tan x = sinx cosx = 1 0 , not defined cot x = cosx sinx = 0 1 = 0 secx = 1 cosx = 1 0 , not defined cosec x = 1 sinx = 1 1 = 1

Prove that : tanA 1 cotA + cotA 1 tanA = 1 + cosecA secA

Marks: 4 Ans

LHS = tanA 1 cotA + cotA 1 tanA = sinA cosA 1 cosA sinA + cosA sinA 1 sinA cosA = sin 2 A cosA sinA cosA + cos 2 A sinA cosA sinA = sin 2 A cosA sinA cosA cos 2 A sinA sinA cosA = sin 3 A cos 3 A sinAcosA sinA cosA = sinA cosA sin 2 A + cos 2 A + sinAcosA sinAcosA sinA cosA = 1 + sinAcosA sinAcosA = 1 + cosecAsecA = RHS Hence proved

The value of sec70°sin20° cos20°cosec70° is

Marks: 1 Ans Ans Not Found in 477722

Without using trigonometric table , evaluate the following : cos 22 ° sin 68 ° + cos sin 90 ° + cosec sec 90 ° 2 tan 1 ° tan 2 ° tan 3 ° . … tan 89 °

We have , cos 22 ° sin 68 ° + cos sin 90 ° + cosec sec 90 ° 2 tan 1 ° tan 2 ° tan 3 ° . … tan 89 ° = cos 90 ° 68 ° sin 68 ° + cos cos + cosec cosec 2 tan 90 ° 89 ° tan 90 ° 88 ° tan 90 ° 87 ° . … tan 89 ° = sin 68 ° sin 68 ° + cos cos + cosec cosec 2 cot 89 ° cot 88 ° cot 87 ° . … tan 45 ° . .. 1 c o t 89 ° = 1 + 1 + 1 2 since , tan = 1 cot , sin 90 ° = cos , tan 90 ° = cot = 1

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Practice as much as you can, especially the revised topics. The more you practice, the better you will get.

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Class 10 Maths Chapter 9 Previous Year Questions - Some Application of Trigonometry

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Q1: If a pole 6 m high casts a shadow 2√3 m long on the ground, then sun's elevation is (a) 60 o (b) 45 o (c) 30 o (d) 90 o [2023, 1 Mark] Ans: (a)

Class 10 Maths Chapter 9 Previous Year Questions - Some Application of Trigonometry

Q2: A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high observes two cars at angles of depression of 30°and 60° which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (Use √3 = 1.73) [2023, 4,5,6 Marks] Ans: Let the tower be CD and points A and B be the positions of two cans on the highway. Height of the tower CD = 75 m. In ΔDCB,

From (i) and (ii). we get BC = 7√3 x √3 = 21m ∴ Height of the tower = 8C + CD = 21 m + 7 m = 28 m

Q1: Two boats are sailing in the sea 80 m apart from each cither towards a cliff AB. The angles of depression of the boats from the top of the cliff are 30 o and 45° respectively, as shown in figure. Find the height of the cliff. [2022, 3 Marks]

Q2: 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, then find the height of the building. [2022, 3 Marks] Ans: Let AB be the tower of height 50m and CD be the building of height h m. Now, in ΔABD,

In ΔABD. we have

Q8: 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°. If the bridge is at a height of 8 m from the banks, then find the width of the river. [2022, 3 Marks]

Q9: The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3 = 1.73) [2022, 4,5,6 Marks] Ans:

Q10: The 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. Ten 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. [2022, 4,5,6 Marks] Ans: Let h be the height of the tower and D be the initial position of car and let DB = a, AB = b

Q11: Case Study: Kite festival Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, shows three kites flying together

In ΔBEC,

(ii) Since, the distance between these two kites is d. ΔABC is a right angle triangle (∵∠ACB = 90°)

Hence, the distance between these two kites is 121.65 m.

Q2: The ratio of the length of a vertical rod and the length of its shadow is 1: √3. Find the angle of elevation of the Sun at that moment. [2020, 1 Mark] Ans: Let AC be the length of vertical rod, AB be the length of its shadow and 0 be the angle of elevation of the sun.

Thus , the height of the pedestal is 2.19 m.

Q3: A man in a boat rowing away from a light house 100 in high takes 2 minutes to change the angle of elevation of the top of the fight house from 60° to 30°. Find the speed of the boat in metres per minute. [Use √3 = 1.732) [2019, 4,5,6 Marks] Ans: Let AB = 100 m be the height of the light house. Let the initial distance be x m and angle is 60°.

Now. after two minutes, new distance be y m and angle is 30°. In ΔABD,

Q5: 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. [2019, 4,5,6 Marks] Ans: Let AB and CD be two poles of height hm. Let P be a point on road such that BP = x so that PD= BD - BP = (80 - x)m

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CBSE 10th Standard Maths Subject Introduction to Trigonometry Case
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 - Complete list of 10th Standard CBSE question papers, syllabus, exam tips, study material, previous year exam question papers, centum ...
Important Questions Class 10 Maths Chapter 8
State whether the following will be true or false. clarify your answer. (i) The value for tan A is always less than 1. (ii) sec A = 12/5 thus some value of the angle A. (iii)cos A is the abbreviation used to the cosecant of the angle A. (iv) cot A is a product of cot A. (v) sin θ = 4/3 for some of the angles θ.
Class 10 Maths Chapter 9 Previous Year Questions
Q11: Case Study: Kite festival ... Introduction of Previous Year Questions: ... The "Previous Year Questions: Some Applications Of Trigonometry Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge their ...

Class 10 Maths Case Study Questions Chapter 8 Introduction to Trigonometry

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