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NCERT Solutions for Exercise 9.1 Class 10 Maths Chapter 9 - Some Applications of Trigonometry

NCERT Solutions for Exercise 9.1 Class 10 Maths Chapter 9 - Some Applications of Trigonometry

Edited By Komal Miglani | Updated on Apr 30, 2025 03:23 PM IST | #CBSE Class 10th

Trigonometry enables people to determine building heights and river distances through measurements when direct approaches are impossible. The exercise transitions to practical trigonometry, which introduces us to elevation and depression angles. To solve real-life problems, we use triangles for visualisation, so we learn about applying sine, cosine and tangent ratios to obtain unknown heights and distances. The education develops both analytical skills and practical usage abilities.

This Story also Contains
  1. NCERT Solutions Class 10 Maths Chapter 9: Exercise 9.1
  2. Access Solution of Some Applications of Trigonometry Class 10 Chapter 9 Exercise: 9.1
  3. Topics Covered in Chapter 9 Applications of Trigonometry: Exercise 9.1
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject-Wise
NCERT Solutions for Exercise 9.1 Class 10 Maths Chapter 9 - Some Applications of Trigonometry
NCERT Solutions for Exercise 9.1 Class 10 Maths Chapter 9 - Some Applications of Trigonometry

Class 10 students should use the NCERT Solutions to solve real-life right-angled triangle problems with basic trigonometric ratios. The exercise teaches students to recognise appropriate triangles, then select proper trigonometric ratios before correctly solving elevation and depression angle problems. The education provided in the exercises provided in NCERT Books serves essential groundwork for advanced trigonometric applications used in navigation, along with surveying and construction work, and astronomy.

NCERT Solutions Class 10 Maths Chapter 9: Exercise 9.1

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Access Solution of Some Applications of Trigonometry Class 10 Chapter 9 Exercise: 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 ACB =30o
Let the height of the pole (AB) be h

So, in the right triangle ΔABC

sdgfhgjhjfdfg4

By using the Sin 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:

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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,
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:

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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, 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 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:

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

In triangle ABQ,

tan30o=hBQ
BQ=h3 .........................(ii)
On equating the eq(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 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=80h3
h3=20
h=203 m
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 question,

In triangle ΔADB ,
tan30o=AB20+x=h20+x13=h20+x20+x=h3x=h320 ............(i)

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

On equating eq (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 , angle of depression of its foot BCE=45o .

According to question,

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

In triangle ΔACE ,

tan60o=hCE=h7=3h=73m

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 from the 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 question,

In triangle ΔADB ,

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

In triangle ΔACB ,

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

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

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 ( 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, distance traveled 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 xms1 . 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 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 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 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 the equation (i) and (ii), we get

h2=36h=6m

Hence the height of the tower is 6 m.

Topics Covered in Chapter 9 Applications of Trigonometry: Exercise 9.1

The questions in exercise 9.1 Class 10 Maths broadly consist of some basic questions in which we have to find height and distance with the help of the formulas of tangent, cosine and sine. Some of the terms used in the problems are as follows.

  • Line of Sight: It is the line drawn from the eye of an observer to the point in the object viewed by the observer.
  • Angle of Elevation: An observer who looks up toward an object will experience this phenomenon. It is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.
    E.g.: Person looking at the top of a tree.
  • Angle of Depression: An observer who looks down toward an object will experience this phenomenon. It is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.
    E.g., standing on a cliff and looking at a boat in the sea.
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The following key sequence directs a solution to these problems:

  1. Spot the right-angled triangle present in the provided problem.
  2. Identify known along with unknown values of height, distance and angle in this problem.
  3. Select the appropriate trigonometric ratio according to the available information.
  4. Solve for the unknown value.
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JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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.

Frequently Asked Questions (FAQs)

1. According to NCERT solutions for Class 10 Maths 9 exercise 9.1 what do you mean by trigonometric ratio.

 The values of all trigonometric functions dependent on the value of the ratio of sides in a right-angled triangle are known as trigonometric ratios.

2. How many topic are covered in NCERT solutions for Class 10 Maths 9 exercise 9.1?
  • angle of depression 

  • angle of elevation 

  • line of sight

  • finding the height or length of an object or the distance between two distant objects

3. What do you mean by angle of elevation according to NCERT solutions for Class 10 Maths 9 exercise 9.1?

The angle created by the line of sight with the horizontal when it is above the horizontal level is the angle of elevation of an item that we can see.

When we elevate our head to gaze at an object, for example.

4. According to NCERT answers for Class 10 Maths 9 exercise 9.1, what do you mean by angle of elevation?

The angle formed by the line of sight with the horizontal when it is below the horizontal level is the angle of depression of an item that we can see.

When we drop our head to gaze at an object, for example.

5. What do you mean by line of sight according to NCERT solutions for Class 10 Maths 9 exercise 9.1?

The line of sight is the line which is a line traced from an observer's eye to a point in the item being seen.

6. An object's height or length, as well as the distance between two distant objects, may be calculated using.

Trigonometric ratios can be used to calculate an object's height or length, as well as the distance between two distant objects.

7. What is the total number of solved examples prior to the Class 10 Maths exercise 9.1?

Before the Class 10 Mathematics chapter 9 activity 9.1, there are seven key questions that must be answered.

8. In the NCERT solutions for Class 10 Maths chapter 9 exercise, how many questions are there?

In the NCERT solutions for the Class 10 Maths chapter 9 exercise, there are 16 problems.

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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.

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3. Counseling and Seat Allocation:

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

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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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