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

Trigonometry is used in real life situations where we want to find angles or distances. For example, in measuring the height of a tall building or a mountain, we can use trigonometric ratios like sine, cosine, or tangent. It’s also used in navigation, which includes locating the location of a ship or a plane. Concepts like line of sight, angle of elevation, and angle of depression help us measure things that are far away.

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1. NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF Free Download
2. Applications Of Trigonometry Class 10 NCERT Solution - Important Notes
3. NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Exercise)
4. NCERT Solutions for Class 10 Maths: Chapter Wise
5. NCERT Exemplar solutions - Subject-wise
6. NCERT Solutions for Class 10 for other subjects
7. NCERT Books and NCERT Syllabus

This article on NCERT Class 10 Maths Chapter 9 Solutions of Some Applications of Trigonometry provides clear and step-by-step solutions for exercise problems in NCERT Class 10 maths book. These NCERT Solutions are created by subject experts at Careers360, considering the latest CBSE syllabus, 2025-26. For further reference, students can check NCERT Notes For Class 10 Mathematics Chapter Some Applications of Trigonometry and NCERT Exemplar Solutions For Class 10 Mathematics Chapter Trigonometry.

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

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Applications Of Trigonometry Class 10 NCERT Solution - Important Notes

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:

• The angle of depression is formed when you have to look down to see something.

• The angle of depression is created between the horizontal line (above the angle) and the line of sight to the object.

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

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).

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

By using the Sin rule
sin⁡θ=PH=ABAC
sin⁡30o=h20
h=10 m.

Hence, the answer is 10 m.

Q2 A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 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:

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 ,
tan⁡θ=PB=x8
tan⁡30o=x8=13
So, the value of x = 8/3

Similarly,
cos⁡30o=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)

Hence, the answer is 13.856 m (approx)

Q3 A contractor plans to install two slides for the children to play in a park. For 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:

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

Given that,
AF = 1.5 m, BC = 3 m, ∠AEF=30o and ∠BDC=60o

In triangle Δ EAF,
sin⁡θ=AFEF=1.5x
sin⁡30o=1.5x
The value of x is 3 m.

Similarly in Δ CDB,
sin⁡60o=3y
32=3y
The value of y is 23 = 2(1.732) = 3.468

Hence the length of the slide for children below 5 years. 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:

Let the height of the tower AB be h and the angle of elevation from the ground at point C is ∠ACB=30o
According to the question,
In the right triangle ΔABC ,
tan⁡θ=ABBC=h30
tan⁡30o=13=h30
The value of h is 103 = 10(1.732) = 17.32 m

Hence, the answer 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:

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,

sin⁡60o=ABAC=60l

32=60l

The value of length of the string ( l ) is 403 = 40(1.732) = 69.28 m

Hence, the answer 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:

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,
⇒tan⁡30o=AFDF=28.5DF⇒13=28.5DF
So, DF = (28.5)3

In right angle triangle ΔAFE ,
tan⁡60o=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:

Suppose BC = h is the height of the transmission tower AB is

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,
tan⁡45o=ABAD=20x
So, x = 20 m

Again,
In triangle Δ CAD,

⇒tan⁡60o=AB+BCAD=20+h20⇒3=1+h20⇒h=20(3−1)⇒20(0.732)=14.64m

Hence, the answer is 14.64m

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:

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 ,
tan⁡45o=1=ABBC=hBC
therefore, BC = h m

In triangle ΔCBD ,
⇒tan⁡60o=BDBC=h+1.6h⇒3=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:

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,
tan⁡60o=PQBQ=50BQ
3=50BQBQ=503 .......................(i)

In triangle ABQ,

tan⁡30o=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 points from the poles.

Answer:

Given that,
The height of both poles is equal to DC = AB. The angle of elevation of the top of the poles is ∠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,

⇒tan⁡30o=DCCE=hx⇒13=hx⇒x=h3 ..............(i)

In triangle AEB,
⇒tan⁡60o=ABBE=h80−x⇒3=h80−x⇒x=80−h3 ..................(ii)
On equating eq (i) and eq (ii), we get

3h=80−h3

h3=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 the 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.

Answer:

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

In triangle ΔADB ,

tan⁡30∘=AB20+x=h20+x⇒13=h20+x⇒20+x=h3⇒x=h3−20

In triangle ACB,
⇒tan⁡60o=hx=3⇒x=h3 .............(ii)

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

h3−20=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:

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

According to the question,

In triangle ΔDBC ,
tan⁡45o=CDBD=7BD=1BD=7 m
since DB = CE = 7 m

In triangle ΔACE ,

tan⁡60o=hCE=h7=3∴h=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 depression angles 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

Let the distance between both the ships be x m.
According to the question,

In triangle ΔADB ,

tan⁡300=ABBD=75x+y=13
∴x+y=753 .............(i)

In triangle ΔACB ,

tan⁡450=1=75BC=75y
∴y=75 m .............(ii)

From equation (i) and (ii) we get;
x=75(3−1)=75(0.732)
x=54.9≃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 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.

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.

Let the x distance traveled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m

Now, In triangle ΔBCA ,

tan⁡600=3=ABBC=87BC∴BC=293

In triangle ΔDCE ,

tan⁡300=13=DECE=87CE∴CE=873

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

=CE−BC=873−293=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:

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 ΔDAC ,
tan⁡300=13=h6x+xtx(6+t)=h3 ..........................(i)

In triangle ΔBCD ,

tan⁡600=3=hxt∴h=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 sec to reach the foot of the tower.

NCERT Exemplar solutions - Subject-wise

To get subject wise exemplars, students can click the following links.

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science

NCERT Solutions for Class 10 for other subjects

To get subject wise NCERT solutions, students can click the following links.

• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Science

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus which is very useful for students before strategizing their study plan. Also, links to some reference books which are important for further studies.

• NCERT Books Class 10 Maths
• NCERT Syllabus Class 10 Maths
• NCERT Books Class 10
• NCERT Syllabus Class 10