Want to know how sailors navigated the oceans or how astronomers measured stars? The secret lies in trigonometry. Applications of trigonometry prove that math is the ladder we climb to measure the surrounding heights. In the last chapter, students learned about trigonometric ratios and identities. This chapter mainly deals with the concepts of heights and distances, angles of elevation, angles of depression, and the line of sight. The primary benefit of using NCERT Solutions for Class 10 is that they offer clear explanations and accurate answers, facilitating a deeper understanding. Many teachers recommend NCERT Solutions because they closely match the exam pattern.
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Heights and distances are an essential part of Trigonometry, as learning these concepts thoroughly will not only help in class 10 but also in future classes and competitive exams. Trigonometry makes heights and distances easy to measure, demonstrating how math can connect the earth with the sky. These NCERT Solutions for Class 10 Maths are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. For details about the syllabus, revisions, and to download PDF files, refer to the NCERT article.
The NCERT Solutions for Class 10 Maths Chapter 9 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.
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NCERT Some Applications of Trigonometry Class 10 Solutions: Exercise 9.1 Total questions: 15 |
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 $\Delta ABC$
By using the sine 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.
Answer:
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)
Answer:
Suppose $x$ m is the length of slides for children below 5 years and the length of slides for elder children is $y$ 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.
Answer:
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 the 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
Answer:
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 the 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.
Answer:
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 the 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 = $19\sqrt{3}$
Answer:
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 the 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
Answer:
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
Answer:
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 the 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 equations (i) and (ii), we get,
$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = 50/3 = 16.66 m = height of the building.
Answer:
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 the 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
$\begin{aligned} & \sqrt{3} h=80-\frac{h}{\sqrt{3}} \\ & \frac{h}{\sqrt{3}}=20\end{aligned}$
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.
Answer:
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 the question,
In triangle $\Delta ADB$ ,
$
\begin{aligned}
& \Rightarrow \tan 30^{\circ}=\frac{A B}{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)
\end{aligned}
$
In triangle ACB,
$\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}$ .............(ii)
On equating equations (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.
Answer:
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 $\angle ACE = 60^o$, the angle of depression of its foot $\angle BCE = 45^o$.
According to the 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 $h+7$ $=7(1+\sqrt{3}) m$
Answer:
Given that,
The height of the lighthouse (AB) is 75 m above sea level. And the angle of depression of two different ships are $\angle ADB = 30^0$ and $\angle ACB = 45^0$ respectively
Let the distance between both the ships be $x$ m.
According to the 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 equations (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 approximately 55 m.
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, $\angle DCE =30^0$.
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, the distance travelled by the balloon from position A to D
$= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3}$ m
Answer:
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 the 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 takes 3 seconds to reach the foot of the tower.
Answer:
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 the 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)
Multiplying equations (i) and (ii), we get
$\\h^2 = 36\\ \Rightarrow h = 6 m$
Hence, the height of the tower is 6 m.
Exercise-wise NCERT Solutions of Some Applications of Trigonometry Class 10 Maths Chapter 9 are provided in the link below.
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Question:
From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
Answer:
We have to find the value of $x$.
In $\triangle ABC,$
$\tan60° = \frac{AC}{BC}$
We know, $CE=BD$
⇒ $AC=AE-CE$
⇒ $AC=57.75-17.75$
⇒ $AC=40$
In $\triangle ABC,$
$\tan60° = \frac{40}{x}$
⇒ $\sqrt3=\frac{40}{x}$
⇒ $x=\frac{40}{\sqrt3}$
⇒ $x=\frac{40\sqrt3}{3}$ m
Hence, the correct answer is $\frac{40\sqrt3}{3}$.
The topics discussed in the NCERT Solutions for class 10 chapter 9 Some Applications of Trigonometry, are:
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.
Case 1:
In this case, we can observe the following:
Height of a tower, hill, or building
Distance of an object from the foot of the tower, hill, or building and sometimes the shadow of it
The angle of elevation or the angle of depression
Any two of the above three parameters will be provided in the question. This type of problem can be solved using the formulas given below.
In the right triangle $ABC$,
$\begin{aligned} & \sin \theta=\frac{\text { Opposite } }{ \text { Hypotenuse }}=\frac{\mathrm{AB} }{ \mathrm{AC}} \\ & \cos \theta=\frac{\text { Adjacent } }{ \text { Hypotenuse }}=\frac{\mathrm{BC} }{ \mathrm{AC}} \\ & \tan \theta=\frac{\text { Opposite } }{ \text { Adjacent }}=\frac{\mathrm{AB} }{\mathrm{BC}}\end{aligned}$
Case 2:
In this case, we can deal with different illustrations. One of the commonly solved problems is about the movement of an observer. If the observer moves toward objects like a tower, building, hill, etc., then the angle of elevation increases. The angle of elevation decreases when the observer moves away from the object. Here, the distance moved by the observer can be found using the formula given below:
In the right triangle given below, $d$ is the distance between $C$ and $D$.
$d = h(\cot x - \cot y)$
Case 3:
There is another case where two different situations happen at the same time. In this case, we get similar triangles with the same angle of elevation or angle of depression. These types of problems can be solved with the help of formulas related to similar triangles.
In the right triangle $ABC, DE || AB,$
Here, triangles $ABC$ and $EDC$ are similar.
Using Thales' or BPT theorem, we can write the ratio of sides as:
$\frac{AB}{ED} = \frac{BC}{DC}$
For students' preparation, Careers360 has gathered all Class 10 Maths NCERT solutions here for quick and convenient access.
Also, read,
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects. The following links will help them in this process.
Students can check the latest syllabus and some reference books from the following links.
Frequently Asked Questions (FAQs)
The angle of elevation is the angle formed when we look up at an object from the ground.
The angle of depression is the angle formed when we look down at an object from a higher point.
In the Some Applications of Trigonometry class 10 questions with solutions, the following important formulas should be remembered:
This article provides NCERT Solutions for Class 10 Maths Chapter 9 PDF. You can just click on it and download. Also, students can download Maths Class 10 Chapter 9 using the official website of Careers360.
Class 10 Chapter 9 of the maths textbook, contains problems on height and distance, solvable using these steps.
The easiest way to understand trigonometric ratios in the NCERT solutions class 10th maths chapter 9 is:
Use these class 10 NCERT Maths Chapter 9 solutions to master these formulas.
On Question asked by student community
Hello,
If you want to get your 10th marksheet online, you just need to visit an official website like https://www.cbse.gov.in/ or https://results.cbse.nic.in/ for the CBSE board, and for the state board, you can check their website and provide your roll number, security PIN provided by the school, and school code to download the result.
I hope it will clear your query!!
Hello, if you are searching for Class 10 books for exam preparation, the right study material can make a big difference. Standard textbooks recommended by the board should be your first priority as they cover the syllabus completely. Along with that, reference books and guides can help in practicing extra questions and understanding concepts in detail. You can check the recommended books for exam preparation from the link I am sharing here.
https://school.careers360.com/ncert/ncert-books-for-class-10
https://school.careers360.com/boards/cbse/cbse-best-reference-books-for-cbse-class-10-exam
Hello
You asked about Class 10 sample paper board exam and most important questions. Practicing sample papers and previous year questions is one of the best ways to prepare for the board exam because it gives a clear idea of the exam pattern and types of questions asked. Schools and teachers usually recommend students to solve at least the last five years question papers along with model papers released by the board.
For Class 10 board exams, the most important areas are Mathematics, Science, Social Science, English, and Hindi or regional language. In Mathematics, questions from Algebra, Linear Equations, Geometry, Trigonometry, Statistics, and Probability are repeatedly seen. For Science, the key chapters are Chemical Reactions, Acids Bases and Salts, Metals and Non metals, Life Processes, Heredity, Light and Electricity. In Social Science, priority should be given to Nationalism, Resources and Development, Agriculture, Power Sharing, Democratic Politics, and Economics related topics. In English, focus on unseen passages, grammar exercises, and important writing tasks like letter writing and essays.
Follow these steps to access the SQPs and marking schemes:
Step 1: Visit https://cbseacademic.nic.in/
Step 2: Click on the link titled “CBSE Sample Papers 2026”
Step 3: A PDF will open with links to Class 10 and 12 sample papers
Step 4: Select your class (Class 10 or Class 12)
Step 5: Choose your subject
Step 6: Download both the sample paper and its marking scheme
If you are looking for Class 10 previous year question papers for 2026 preparation, you can easily access them through the links I’ll be attaching. These papers are very helpful because they give you a clear idea about the exam pattern, marking scheme, and the type of questions usually asked in board exams. Practicing these will not only improve your time management but also help you identify important chapters and commonly repeated questions.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-10
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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In the reaction,
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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