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

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

Edited By Komal Miglani | Updated on Jun 28, 2025 08:47 AM IST | #CBSE Class 10th
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CBSE Class 10th  Exam Date : 10 Jul' 2025 - 15 Jul' 2025

Trigonometry isn’t just about triangles; it’s about seeing the world through angles and heights. 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. The NCERT Solutions for Class 10 Maths Chapter 9, Some Applications of Trigonometry, explain the daily uses of trigonometry to students. This chapter mainly deals with the concepts of heights and distances, angles of elevation, angles of depression, and the line of sight.

This Story also Contains
  1. NCERT Solution for Class 10 Maths Chapter 9 Solutions: Download PDF
  2. NCERT Solutions for Class 10 Maths Chapter 9: Exercise Questions
  3. Some Applications of Trigonometry Class 10 Chapter 9: Topics
  4. Some Applications of Trigonometry Class 10 Solutions: Important Formulae
  5. NCERT Solutions for Class 10 Mathematics - Chapter Wise
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Trigonometry makes heights and distances easy to measure, demonstrating how math can connect the earth with the sky. The NCERT solutions for Class 10 presented here are based on the revised CBSE syllabus for 2025-26, available from Careers360 subject matter experts who wrote these easy-to-understand solutions and showed a stepwise explanation of each question with connected pictures. For details about the syllabus, revisions, and to download PDF files, refer to this link: NCERT.

NCERT Solution for Class 10 Maths Chapter 9 Solutions: Download PDF

Download PDF


NCERT Solutions for Class 10 Maths Chapter 9: Exercise Questions

NCERT Some Applications of Trigonometry Class 10 Solutions: Exercise 9.1
Page number: 141-143

Total questions: 15

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 ACB =30o
Let the height of the pole (AB) be h

So, in the right triangle ΔABC

sdgfhgjhjfdfg4

By using the sine rule
sinθ=PH=ABAC
sin30o=h20
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:

1594291520636
Suppose DB is a tree, and the AD is the broken height of the tree, which touches the ground at C.
Given that,
ACB=30o , BC = 8 m
let AB = x m and AD = y m
So, AD+AB = DB = x+y

In right angle triangle ΔABC ,
tanθ=PB=x8
tan30o=x8=13
So, the value of x = 8/3

Similarily,
cos30o=BCAC=8y
the value of y is 16/3

So, the total height of the tree is-

x+y=243=83

= 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:

1594290251569
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, AEF=30o and BDC=60o

In triangle Δ EAF,
sinθ=AFEF=1.5x
sin30o=1.5x
The value of x is 3 m.

Similarily in Δ CDB,
sin60o=3y
32=3y
the value of y is 23 = 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 ACB=30o
According to the question,
In the right triangle ΔABC ,
tanθ=ABBC=h30
tan30o=13=h30
the value of h is 103 = 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:
1594290308744
Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is ACB=60o.
Let the length of the string AC be l.
According to the question,
In right triangle Δ CBA,
sin60o=ABAC=60l
32=60l
The value of length of the string ( l ) is 403 = 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:

1594290331910
Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
ADF=30o and AEF=60o

According to the question,
In right triangle AFD,
tan30o=AFDF=28.5DF13=28.5DF
So, DF = (28.5)3

In right angle triangle ΔAFE ,
tan60o=AFFE=28.5EF
3=28.5EF
EF = 9.53

So, distance walked by the boy towards the building = DF - EF = 193

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
CDA=60o and BDA=45o

According to the question,
In triangle Δ BDA,
tan45o=ABAD=20x
So, x = 20 m

Again,
In triangle Δ CAD,

tan60o=AB+BCAD=20+h203=1+h20h=20(31)20(0.732)=14.64m

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:

1594290373464
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( DCB= 60o )and( ACB= 45o ) respectively.

Now,
In triangle ΔABC ,
tan45o=1=ABBC=hBC
therefore, BC = h m

In triangle ΔCBD ,
tan60o=BDBC=h+1.6h3=1+1.6h
the value of h is 0.8(3+1) m
Hence the height of the pedestal is 0.8(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. AQB=30o and PBQ=60o
Let the height of the building be h m

According to the question,
In triangle PBQ,
tan60o=PQBQ=50BQ
3=50BQBQ=503 .......................(i)

In triangle ABQ,

tan30o=hBQ
BQ=h3 .........................(ii)
On equating equations (i) and (ii), we get,

503=h3
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 DEC=30o and AEB=60o resp.
Let the height of the poles be h m and CE = x and BE = 80 - x

According to the question,
In triangle DEC,

tan30o=DCCE=hx13=hxx=h3 ..............(i)

In triangle AEB,
tan60o=ABBE=h80x3=h80xx=80h3 ..................(ii)
On equating eq (i) and eq (ii), we get

3h=80h3h3=20
So, x = 60 m

Hence, the height of both poles is ( h=203 )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 the question,

In triangle ΔADB ,

tan30=AB20+x=h20+x13=h20+x20+x=h3x=h320(i)

In triangle ACB,
tan60o=hx=3x=h3 .............(ii)

On equating equations (i) and (ii), we get:

h320=h3
from here we can calculate the value of h=103=10(1.732)=17.32m 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:

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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 ACE=60o, the angle of depression of its foot BCE=45o .

According to the question,

In triangle ΔDBC ,
tan45o=CDBD=7BD=1BD=7 m
since DB = CE = 7 m

In triangle ΔACE ,

tan60o=hCE=h7=3h=73 m

Thus, the total height of the tower equal to h+7 =7(1+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 above sea level. And the angle of depression of two different ships are ADB=300 and ACB=450 respectively
1594290544799
Let the distance between both the ships be x m.
According to the question,

In triangle ΔADB ,

tan300=ABBD=75x+y=13
x+y=753 .............(i)

In triangle ΔACB ,

tan450=1=75BC=75y
y=75 m .............(ii)

From equations (i) and (ii) we get;
x=75(31)=75(0.732)
x=54.955 m

Hence, the distance between the two ships is approximately 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 travelled 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 ( ACB=600 ), and after some time, DCE=300.
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 ΔBCA ,

tan600=3=ABBC=87BCBC=293

In triangle ΔDCE ,

tan300=13=DECE=87CECE=873

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

=CEBC=873293=583 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:

1594290668622
Let h be the height of the tower (DC) and the speed of the car be x ms1 . 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 ΔDAC ,
tan300=13=h6x+xtx(6+t)=h3 ..........................(i)

In triangle ΔBCD ,

tan600=3=hxth=3.xt ...................(ii)

Put the value of h in equation (i) we get,
x(6+t)=(3.3)xt6x+xt=3xt6x=2xt
t=3

Hence, from point B car takes 3 seconds 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 BQA=θ , so APB=90θ

According to question,

In triangle ΔABQ ,

tanθ=h9h=9tanθ ..............(i)

In triangle ΔABP ,

tan(90θ)=cotθ=h4h=4cotθ .....................(ii)

Multiply equations (i) and (ii), we get

h2=36h=6m

Hence, the height of the tower is 6 m.

Some Applications of Trigonometry Class 10 Chapter 9: Topics

The topics discussed in the NCERT Solutions for class 10, chapter 9, Some Applications of Trigonometry, are:

  • Heights and Distances
  • Summary
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Some Applications of Trigonometry Class 10 Solutions: Important Formulae

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,
sinθ= Opposite  Hypotenuse =ABACcosθ= Adjacent  Hypotenuse =BCACtanθ= Opposite  Adjacent =ABBC

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(cotxcoty)

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:
ABED=BCDC

NCERT Solutions for Class 10 Mathematics - Chapter Wise

For students' preparation, Careers360 has gathered all Class 10 Maths NCERT solutions here for quick and convenient access.

Also, read,

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject-Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

NCERT Books and NCERT Syllabus

Students can check the latest CBSE syllabus from the following links.

Frequently Asked Questions (FAQs)

1. What are the important topics covered in Class 10 Maths Chapter 9 in ncert?

In this chapter, Trigonometry applications class 10 covered the following topics-

1. Introduction 

2. Line of Sight

3. Angle of Elevation

4. Angle of Depression

5. Trigonometric Ratios in Heights and Distances

6. Solving Problems Based on Heights and Distances

7. Assumptions in Height and Distance Problems

2. How can I download the NCERT Solutions for Class 10 Maths Chapter 9 PDF?

This article provides NCERT Solutions for Class 10 Maths Chapter 9 PDF. You can just click and download it. Also, students can download Maths Class 10, chapter 9 using the official website of Careers360.

3. How to solve height and distance problems in Class 10 Maths?

Class 10, chapter 9 of the math textbook, contains problems on height and distance, solvable using these steps.

  • Draw a Diagram: Convert the given problem into a right-angled triangle.
  • Find angles (elevation/depression), heights, or distances.
  • Choose the Right Trigonometric Ratio:
    • tan θ = Perpendicular / Base
    • sin θ = Perpendicular / Hypotenuse
    • cos θ = Base / Hypotenuse
  • Apply the Formula: Use the best trigonometric ratio.
  • Solve the Equation: Find unknown values grade by grade.
4. What is the easiest way to understand trigonometric ratios in Chapter 9?

The easiest way to understand trigonometric ratios in NCERT solutions class 10th maths chapter 9 is:

  1. Learn the Basic Ratios:
    • sin θ = Perpendicular / Hypotenuse
    • cos θ = Base / Hypotenuse
    • tan θ = Perpendicular / Base
  2. Use the Mnemonic: SOH-CAH-TOA
    • Sin = Opposite/Hypotenuse
    • Cos = Adjacent/Hypotenuse
    • Tan = Opposite/Adjacent
  3. Practice Right-Angled Triangle Problems
     
  4. Relate to Real-Life Applications (heights, distances, etc.)

Use these class 10 NCERT Maths Chapter 9 solutions to master these formulas.

5. What are the important formulas to remember in Some Applications of Trigonometry?

In Applications of trigonometry class 10 questions with solutions, the following important formulas should be remembered:

Basic Trigonometric Ratios:

  1. sin θ = Perpendicular / Hypotenuse
  2. cos θ = Base / Hypotenuse
  3. tan θ = Perpendicular / Base
  4. cosec θ = 1 / sin θ
  5. sec θ = 1 / cos θ
  6. cot θ = 1 / tan θ

Height and Distance Formulas:

  1. Tan θ = Height / Distance (for the angle of elevation/depression)
  2. The angle of Elevation: The observer looks upward
  3. Angle of Depression: Observer looks downward

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Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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