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

Edited By Dinesh | Updated on Feb 8, 2023 - 1:30 p.m. IST | #CBSE Class 10th

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. 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.
NCERT solutions for Class 10 Maths chapter 9 are designed by the subject experts to help students in their preparation of board exams. Also, check NCERT solutions for Class 6 to 12.

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

Also Refer,

NCERT Solutions for Class 10 Maths Chapter 9 - All Exercises

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Excercise: 9.1

Q1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

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Answer:

Given that,
The length of the rope (AC) = 20 m. and $\angle ACB$ $=30^o$
Let the height of the pole (AB) be $h$

So, in the right triangle $\Delta ABC$

sdgfhgjhjfdfg4

By using the Sin rule
$\sin \theta = \frac{P}{H} =\frac{AB}{AC}$
$\sin 30^o =\frac{h}{20}$
$h =10$ m.
Hence the height of the pole is 10 m.

Q2 A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer:

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Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
$\angle ACB = 30^o$ , BC = 8 m
let AB = $x$ m and AD = $y$ m
So, AD+AB = DB = $x+y$

In right angle triangle $\Delta ABC$ ,
$\tan \theta = \frac{P}{B}=\frac{x}{8}$
$\tan 30^o =\frac{x}{8}=\frac{1}{\sqrt{3}}$
So, the value of $x$ = $8/\sqrt{3}$

Similarily,
$\cos 30^o = \frac{BC}{AC} = \frac{8}{y}$
the value of $y$ is $16/\sqrt{3}$

So, the total height of the tree is-

$x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}$

= 8 (1.732) = 13.856 m (approx)

Q3 A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer:

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Suppose $x$ m is the length of slides for children below 5 years and the length of slides for elders children be $y$ m.

Given that,
AF = 1.5 m, BC = 3 m, $\angle AEF = 30^o$ and $\angle BDC = 60^o$

In triangle $\Delta$ EAF,
$\sin \theta = \frac{AF}{EF} = \frac{1.5}{x}$
$\sin 30^o = \frac{1.5}{x}$
The value of $x$ is 3 m.

Similarily in $\Delta$ CDB,
$\sin 60^o = \frac{3}{y}$
$\frac{\sqrt{3}}{2}= \frac{3}{y}$
the value of $y$ is $2\sqrt{3}$ = 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.

Q4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer:

1594290279361
Let the height of the tower AB is $h$ and the angle of elevation from the ground at point C is $\angle ACB = 30^o$
According to question,
In the right triangle $\Delta ABC$ ,
$\tan \theta = \frac{AB}{BC} = \frac{h}{30}$
$\tan 30^o =\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

Q5 A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

A
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Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is $\angle ACB = 60^o$ .
Let the length of the string AC be $l$ .
According to question,
In right triangle $\Delta$ CBA,
$\sin 60^{o} = \frac{AB}{AC} = \frac{60}{l}$
$\frac{\sqrt{3}}{2} = \frac{60}{l}$
The value of length of the string ( $l$ ) is $40\sqrt{3}$ = 40(1.732) = 69.28 m
Hence the length of the string is 69.28 m.

Q6 A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

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Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
$\angle ADF = 30^o$ and $\angle AEF = 60^o$

According to question,
In right triangle AFD,
$\\\Rightarrow \tan 30^o=\frac{AF}{DF} = \frac{28.5}{DF}\\\\\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}$
So, DF = $(28.5)\sqrt{3}$

In right angle triangle $\Delta AFE$ ,
$\tan 60^o =\frac{AF}{FE}=\frac{28.5}{EF}$
$\sqrt{3}=\frac{28.5}{EF}$
EF = $9.5\sqrt{3}$

So, distance walked by the boy towards the building = DF - EF = $19\sqrt{3}$

Q7 From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer:

1594290351178
Suppose BC = $h$ 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 = $h$ m and AD = $x$ m
$\angle CDA = 60^o$ and $\angle BDA = 45^o$

According to question,
In triangle $\Delta$ BDA,
$\tan 45^o = \frac{AB}{AD}=\frac{20}{x}$
So, $x$ = 20 m

Again,
In triangle $\Delta$ CAD,

$\\\Rightarrow \tan 60^o = \frac{AB+BC}{AD}=\frac{20+h}{20}\\\\\Rightarrow \sqrt{3}= 1+\frac{h}{20}\\\\\Rightarrow h=20(\sqrt{3}-1)\\\\\Rightarrow 20(0.732) = 14.64 m$

Answer- the height of the tower is 14.64 m

Q8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer:

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Let the height of the pedestal be $h$ 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( $\angle DCB=$ $60^o$ )and( $\angle ACB=$ $45^o$ ) respectively.

Now,
In triangle $\Delta ABC$ ,
$\tan 45^o =1 =\frac{AB}{BC}=\frac{h}{BC}$
therefore, BC = $h$ m

In triangle $\Delta CBD$ ,
$\\\Rightarrow \tan 60^o = \frac{BD}{BC}=\frac{h+1.6}{h}\\\\\Rightarrow \sqrt{3}= 1+\frac{1.6}{h}$
the value of $h$ is $0.8(\sqrt{3}+1)$ m
Hence the height of the pedestal is $0.8(\sqrt{3}+1)$ m

Q9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer:

1594290421503
It is given that, the height of the tower (AB) is 50 m. $\angle AQB = 30^o$ and $\angle PBQ = 60^o$
Let the height of the building be $h$ m

According to question,
In triangle PBQ,
$\tan 60^o = \frac{PQ}{BQ} = \frac{50}{BQ}$
$\\\sqrt{3} = \frac{50}{BQ}\\ BQ = \frac{50}{\sqrt{3}}$ .......................(i)

In triangle ABQ,

$\tan 30^o = \frac{h}{BQ}$
${BQ}=h\sqrt{3}$ .........................(ii)
On equating the eq(i) and (ii) we get,

$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = 50/3 = 16.66 m = height of the building.

Q10 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.

Answer:

1594290440253
Given that,
The height of both poles are equal DC = AB. The angle of elevation of of the top of the poles are $\angle DEC=30^o$ and $\angle AEB=60^o$ resp.
Let the height of the poles be $h$ m and CE = $x$ and BE = 80 - $x$

According to question,
In triangle DEC,

$\\\Rightarrow \tan 30^o = \frac{DC}{CE} = \frac{h}{x}\\\\\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}\\\\\Rightarrow x=h\sqrt{3}$ ..............(i)

In triangle AEB,
$\\\Rightarrow \tan 60^o = \frac{AB}{BE}=\frac{h}{80-x}\\\\\Rightarrow \sqrt{3}=\frac{h}{80-x}\\\\\Rightarrow x=80 - \frac{h}{\sqrt{3}}$ ..................(ii)
On equating eq (i) and eq (ii), we get

$\sqrt{3}h=80 - \frac{h}{\sqrt{3}}$

$\frac{h}{\sqrt{3}}=20$ So, $x$ = 60 m

Hence the height of both poles is ( $h=20\sqrt{3}$ )m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.

Q11 A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

1594290461635

Answer:

1594290476491
Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that, the width of CD is 20 m,
According to question,

In triangle $\Delta ADB$ ,
$\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20$ ............(i)

In triangle ACB,
$\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}$ .............(ii)

On equating eq (i) and (ii) we get:

$h\sqrt{3}-20= \frac{h}{\sqrt{3}}$
from here we can calculate the value of $h=10\sqrt{3}= 10 (1.732) = 17.32\: m$ and the width of the canal is 10 m.

Q12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer:

1594290494703
Let the height of the cable tower be (AB = $h+7$ )m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower $\angle ACE = 60^o$ , angle of depression of its foot $\angle BCE = 45^o$ .

According to question,

In triangle $\Delta DBC$ ,
$\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m

In triangle $\Delta ACE$ ,

$\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}\) m$

Q13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

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 $\angle ADB = 30^0$ and $\angle ACB = 45^0$ respectively
1594290544799
Let the distance between both the ships be $x$ m.
According to question,

In triangle $\Delta ADB$ ,

$\tan 30^0 = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}$
$\therefore x+y = 75 \sqrt{3}$ .............(i)

In triangle $\Delta ACB$ ,

$\tan 45^0 = 1 =\frac{75}{BC}=\frac{75}{y}$
$\therefore y =75\ m$ .............(ii)

From equation (i) and (ii) we get;
$x = 75(\sqrt{3}-1)=75(0.732)$
$x = 54.9\simeq 55\ m$

Hence, the distance between the two ships is approx 55 m.

Q14 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance traveled by the balloon during the interval.

1594290601110

Answer:

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 ( $\angle ACB =60^0$ ) and after some time $\angle DCE =30^0$ .
1594290642418
Let the $x$ 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 $\Delta BCA$ ,

$\\\tan 60^0 = \sqrt{3}=\frac{AB}{BC}=\frac{87}{BC}\\ \therefore BC = 29\sqrt{3}$

In triangle $\Delta DCE$ ,

$\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE}\\ \therefore CE = 87\sqrt{3}$

Thus, distance traveled by the balloon from position A to D

$= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3}$ m

Q15 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

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Let $h$ be the height of the tower (DC) and the speed of the car be $x\ ms^{-1}$ . Therefore, the distance (AB)covered by the car in 6 seconds is 6 $x$ m. Let $t$ time required to reach the foot of the tower. So, BC = $x$ $t$

According to question,
In triangle $\Delta DAC$ ,
$\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{h}{6x+xt}\\ x(6+t) = h\sqrt{3}$ ..........................(i)

In triangle $\Delta BCD$ ,

$\\\tan 60^0 = \sqrt{3} = \frac{h}{xt}\\ \therefore h = 3.xt$ ...................(ii)

Put the value of $h$ in equation (i) we get,
$\\x(6+t) = (\sqrt{3}.\sqrt{3})xt\\ 6x +xt = 3xt\\ 6x = 2xt$
$t = 3$

Hence, from point B car take 3 sec to reach the foot of the tower.

Q16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer:

1594290687445
Let the height of the tower be $h$ m.
we have PB = 4m and QB = 9 m
Suppose $\angle BQA = \theta$ , so $\angle APB =90- \theta$

According to question,

In triangle $\Delta ABQ$ ,

$\\\tan \theta = \frac{h}{9}\\ \therefore h = 9 \tan \theta$ ..............(i)

In triangle $\Delta ABP$ ,

$\\\tan (90-\theta)=\cot \theta = \frac{h}{4}\\ \therefore h = 4\cot \theta$ .....................(ii)

multiply the equation (i) and (ii), we get

$\\h^2 = 36\\ \Rightarrow h = 6 m$

Hence the height of the tower is 6 m.

NCERT Solutions for Class 10 Maths for Other Chapters

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	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

Subject wise NCERT Exemplar solutions

Know more about NCERT Solutions for Class 10 Maths Chapter 9

As per NCERT Application of Trigonometry Class 10, Trigonometry is one of the most ancient topics studied by scholars all over the world. Ch 9 Maths Class 10 introduces new terms like the line of sight, angle of depression and angle of elevation. NCERT solutions for Class 10 Maths chapter 9 will give you assistance while practising the questions of the exercises 9.1.

The main topics covered in the application of trigonometry class 10

9.1 Introduction: This section 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 topic 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 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 elevation

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.

The angle of depression

How to use NCERT Solutions for Class 10 Maths Chapter 9?

Before coming to this Class 10 Chapter 9, 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 NCERT solutions for Class 10 Maths chapter 9 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.

Keep working hard & happy learning!

NCERT Solutions for Class 10 for other subjects.

Also Check NCERT Books and NCERT Syllabus here

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Question: What are the benefits of studying the class 10 maths applications of trigonometry solutions?

Answer:

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.

Question: How does the NCERT solutions are helpful in CBSE board exam?

Answer:

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.

Question: Are the NCERT Solutions for Class 10 Maths Chapter 9 sufficient for CBSE ex

Answer:

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.

Question: Where can one find the applications of trigonometry class 10 ncert solutions?

Answer:

Some applications of trigonometry class 10 ncert 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.

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Questions related to CBSE Class 10th

Showing 71 out of 71 Questions

403 Views

my sons dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

Shubhangi 30th Jul, 2022

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

1467 Views

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

Aditi Singh 22nd Jul, 2022

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

132 Views

i got less marks in maths can it get replaced by additional hindi in cbse

Aditi Singh 22nd Jul, 2022

Replacement of marks of additional subjects and your main course subject is not like they will swap the marks you got in Hindi and Mathematics on your marksheet. It works like if you calculate the total percentage you got in your class, you can take the marks of your additional subject to consider in place of the subject you have got the least marks in. CBSE schools consider this method only to release the merit list of their students.

If you're in 10th and aiming to get admission in 11th in any school, it depends on the schools if they consider your main five subject marks or the best five subject marks for the admission.

If you're in 12th and are trying for different colleges to get in, it depends on your course which you wish to study further and also some criterias and eligibility of the colleges. In most cases for general courses, colleges take the best three or best four subjects marks consideration for the admission process.

So, good luck.

93 Views

when will cbse term 2 result will declare please tale

Shubham Dhane 21st Jul, 2022

Hello,

Central Board of Secondary Education will declare term 2 CBSE exam result 2022 for Class 10 in the third week of July. Please keep an eye on the official website or the link below for the latest information. Students need to enter their CBSE board roll number and other details to check term 2 cbseresults.nic.in 2022 Class 10 results. The board had announced the CBSE Class 10 results 2022 for term 1 on March 11, 2022.

https://school.careers360.com/boards/cbse/cbse-result