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NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry is the 9th chapter of Maths which deals with the applications of trigonometry in our daily life. Chapter 9 class 10 Maths NCERT solutions are provided here. These NCERT solutions are created by subjects experts at Careers360 keeping in mind latest syllabus of CBSE 2023-24. NCERT Class 10 maths solutions for chapter 9 consist of the answers to each question given in the 9th chapter of NCERT class 10 maths book.
There is only one exercise in NCERT chapter 9 Class 10 Maths which contain 16 questions. In these NCERT solutions for Class 10 Maths chapter 9, students will learn how trigonometry is used for finding the distances and heights and various objects, without actually measuring them. Also, check NCERT solutions for Class 6 to 12.
Also Refer,
Line of Sight - The Line of Sight is the line formed by our vision as it passes through an item when we look at it.
Horizontal Line - The distance between the observer and the object is measured by a horizontal line.
The angle of Elevation:
The angle formed by the line of sight to the top of the item and the horizontal line is called an angle of elevation.
It is above the horizontal line, i.e., when we gaze up at the item, we make an angle of elevation.
The angle of Depression:
When the spectator must look down to perceive the item, an angle of depression is formed.
When the horizontal line is above the angle, the angle of depression is formed between it and the line of sight.
Free download NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF for CBSE Exam.
NCERT Solutions for ch 9 maths class 10 Excercise: 9.1
Given that,
The length of the rope (AC) = 20 m. and
Let the height of the pole (AB) be
So, in the right triangle
By using the Sin rule
m.
Hence the height of the pole is 10 m.
Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
, BC = 8 m
let AB = m and AD =
m
So, AD+AB = DB =
In right angle triangle ,
So, the value of =
Similarily,
the value of is
So, the total height of the tree is-
= 8 (1.732) = 13.856 m (approx)
Suppose m is the length of slides for children below 5 years and the length of slides for elders children be
m.
Given that,
AF = 1.5 m, BC = 3 m, and
In triangle EAF,
The value of is 3 m.
Similarily in CDB,
the value of is
= 2(1.732) = 3.468
Hence the length of the slide for children below 5 yrs. is 3 m and for the elder children is 3.468 m.
Let the height of the tower AB is and the angle of elevation from the ground at point C is
According to question,
In the right triangle ,
the value of is
= 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m
A
Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is .
Let the length of the string AC be .
According to question,
In right triangle CBA,
The value of length of the string ( ) is
= 40(1.732) = 69.28 m
Hence the length of the string is 69.28 m.
Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
and
According to question,
In right triangle AFD,
So, DF =
In right angle triangle ,
EF =
So, distance walked by the boy towards the building = DF - EF =
Suppose BC = is the height of transmission tower and the AB be the height of the building and AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = m and AD =
m
and
According to question,
In triangle BDA,
So, = 20 m
Again,
In triangle CAD,
Answer- the height of the tower is 14.64 m
Let the height of the pedestal be m. and the height of the statue is 1.6 m.
the angle of elevation of the top of the statue and top of the pedestal is(
)and(
) respectively.
Now,
In triangle ,
therefore, BC = m
In triangle ,
the value of is
m
Hence the height of the pedestal is m
It is given that, the height of the tower (AB) is 50 m. and
Let the height of the building be m
According to question,
In triangle PBQ,
.......................(i)
In triangle ABQ,
.........................(ii)
On equating the eq(i) and (ii) we get,
therefore, = 50/3 = 16.66 m = height of the building.
Given that,
The height of both poles are equal DC = AB. The angle of elevation of of the top of the poles are and
resp.
Let the height of the poles be m and CE =
and BE = 80 -
According to question,
In triangle DEC,
..............(i)
In triangle AEB,
..................(ii)
On equating eq (i) and eq (ii), we get
So,
= 60 m
Hence the height of both poles is ( )m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.
Suppose the is the height of the tower AB and BC =
m
It is given that, the width of CD is 20 m,
According to question,
In triangle ,
............(i)
In triangle ACB,
.............(ii)
On equating eq (i) and (ii) we get:
from here we can calculate the value of and the width of the canal is 10 m.
Let the height of the cable tower be (AB = )m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower , angle of depression of its foot
.
According to question,
In triangle ,
since DB = CE = 7 m
In triangle ,
Thus, the total height of the tower equal to
Given that,
The height of the lighthouse (AB) is 75 m from the sea level. And the angle of depression of two different ships are and
respectively
Let the distance between both the ships be m.
According to question,
In triangle ,
.............(i)
In triangle ,
.............(ii)
From equation (i) and (ii) we get;
Hence, the distance between the two ships is approx 55 m.
Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m and the angle of elevation of the balloon from the eye of the girl at any instant is ( ) and after some time
.
Let the distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m
Now, In triangle ,
In triangle ,
Thus, distance traveled by the balloon from position A to D
m
Let be the height of the tower (DC) and the speed of the car be
. Therefore, the distance (AB)covered by the car in 6 seconds is 6
m. Let
time required to reach the foot of the tower. So, BC =
According to question,
In triangle ,
..........................(i)
In triangle ,
...................(ii)
Put the value of in equation (i) we get,
Hence, from point B car take 3 sec to reach the foot of the tower.
Let the height of the tower be m.
we have PB = 4m and QB = 9 m
Suppose , so
According to question,
In triangle ,
..............(i)
In triangle ,
.....................(ii)
multiply the equation (i) and (ii), we get
Hence the height of the tower is 6 m.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
9.1 Introduction: This section of ch 9 maths class 10 introduces the concept of how trigonometry is used in everyday life, and how it was first developed for use in astronomy. The chapter will also look at how trigonometry can be used to find heights and distances without directly measuring them.
9.2 Heights and Distances: this ch 9 maths class 10 delves into the concepts of line of sight, angle of deviation, angle of elevation, and angle of depression, and demonstrates how to solve problems using trigonometric ratios.
9.3 Summary: This section of chapter 9 class 10 maths provides a summary of the key points covered in the chapter, helping students to focus on the most important concepts.
Let's understand a few terms used in Some Applications of Trigonometry Class 10 Chapter through the diagram.
The line of sight- It is an imaginary line drawn from the observer's eye to the object viewed by the observer.
The angle of elevation- The angle of elevation of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is above the horizontal level,
The angle of depression- The angle of depression of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is below the horizontal level.
Before coming to this ch 9 class 10 maths, please ensure that you have completed the previous chapter Introduction to Trigonometry.
Read the conceptual theory given in the NCERT textbook and have a look over some examples present in the textbook for the deeper understanding of the topics.
Once you complete the above points then you can jump to the practice exercises available in the NCERT book.
While practicing the questions in practice exercises, you can use ch 9 class 10 maths solutions to know how to solve, what is the final answer and so on.
When you have done the practice exercise then the best thing to make your concepts strong is the last 5 years CBSE class 10 question papers and the NCERT exemplar problems.
ch 9 maths class 10 examples solutions can be downloaded for free in PDF format from the Craerrs360 website. These NCERT solutions can also find above in this article. Students can practice them to score well in the exam. Also, students can download the application of trigonometry pdf format to study offline.
It is important to thoroughly study and practice all the questions in the NCERT Solutions for trigonometry applications class 10. Once you have completed all the questions, you can supplement your learning by consulting other reference materials and the questions provided in the NCERT Exemplar textbooks.
In CBSE board exam most of the questions are directly asked from NCERT textbook, so students must know topics in the NCERT syllabus very well. In NCERT class 10 maths book. , you will get are detailed solutions provided by the experts who knows how best to answer in the board exam in order to get good marks. Students can practice more questions from NCERT Exemplar.
Some of the benefits of studying NCERT Solutions for application to trigonometry class 10 include the ability to calculate heights and distances of objects without direct measurement and a better understanding of trigonometry-based questions. This knowledge will aid students in answering questions related to trigonometry and performing well in class tests and CBSE exams.
Exam Date:01 January,2025 - 14 February,2025
Exam Date:01 January,2025 - 14 February,2025
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