NCERT SOLUTIONS {2023-2024}
CLASS X MATHEMATICS
SOME APPLICATIONS OF TRIGONOMETRY
CHAPTER - 9 (EXCERCISE 9.1)
QUESTION 1
A circus artist is climbing a 20cm long rope, which is tightly stretched and tied from the top
of vertical pole to the ground .Find the height of the pole, if the angle made by the rope with the
ground level is 30 degree?(CBSE-CLASSX -NCERT)
ANSWER
It can be observed from the figure that AB is the pole
In triangle ABC,
AB/BC = Sin 30
AB/20 =1/2
AB =20/2 =10
Therefore ,the height of the pole is 10 m
QUESTION 2
A tree breaks due to storm and the broken parts bends so that the top of the tree touches the
ground making an angle 30 degree with it .The distance between the foot of the tree to the point
where the top touches the ground is 8m.Find the height of the tree?
ANSWER
BC = 8 m
To Find: Height of the tree, which is AB
From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC
In right ΔABC,
cos 30° = BC/AC
cos 30° = √3/2
3/2 = 8/AC
AC = 16/√3 …(1)
Also,
tan 30° = AB/BC
1/√3 = AB/8
AB = 8/√3 ….(2)
Therefore, total height of the tree = AB + AC
= 16/√3 + 8/√3
= 24/√3
= 8√3 m.
QUESTION 3
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 of3m, and inclined at an angle of 60° to the ground. What should
be the length of the slide in each case?
ANSWER
Let, ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
To Find: AC and PR
In right ΔABC,
sin 30° = AB/AC
1/2 = 1.5/AC
AC = 3
In right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3
Hence, length of the slide for below 5 = 3 m and
Length of the slide for elders children = 2√3 m
QUESTION 4
4. 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 AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the
tower.
To Find: AB (height of the tower)
In right ABC
tan 30° = AB/BC
1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.
QUESTION 5
ANSWER
Draw a figure, based on given instruction,
Let BC = Height of the kite from the ground, BC = 60 m
AC = Inclined length of the string from the ground and
A is the point where string of the kite is tied.
To Find: Length of the string from the ground i.e. the value of AC
From the above figure,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.
QUESTION 6
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
Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the point X
with inclination 60°.
To Find: The distance boy walked towards the building i.e. XY
From figure,
XY = CD.
Height of the building = AZ = 30 m.
AB = AZ – BZ
= 30 – 1.5 = 28.5
Measure of AB is 28.5 m
In right ΔABD,
tan 30° = AB/BD
1/√3 = 28.5/BD
BD = 28.5√3 m
Again,
In right ΔABC,
tan 60° = AB/BC
√3 = 28.5/BC
BC = 28.5/√3 = 28.5√3/3
Therefore, the length of BC is 28.5√3/3 m.
XY = CD = BD – BC
= (28.5√3-28.5√3/3)
= 28.5√3(1-1/3)
= 28.5√3 × 2/3
= 57/√3 = 19√3 m.
Thus, the distance boy walked towards the building is 19√3 m.
QUESTION 7
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
Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC – BC
To Find: AB, Height of the tower
From figure,
In right ΔBCD,
tan 45° = BC/CD
1 = 20/CD
CD = 20
Again,
In right ΔACD,
tan 60° = AC/CD
√3 = AC/20
AC = 20√3
Now, AB = AC – BC = (20√3-20) = 20(√3-1)
Height of transmission tower = 20(√3 – 1) m.
QUESTION 8
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 AB be the height of statue.
D is the point on the ground from where the elevation is taken.
To Find: Height of pedestal = BC = AC-AB
From figure,
In right triangle BCD,
tan 45° = BC/CD
1 = BC/CD
BC = CD …..(1)
Again,
In right ΔACD,
tan 60° = AC/CD
√3 = ( AB+BC)/CD
√3CD = 1.6 + BC
√3BC = 1.6 + BC (using equation (1)
√3BC – BC = 1.6
BC(√3-1) = 1.6
BC = [(1.6)(√3+1)]/[(√3-1)(√3+1)]
BC = [1.6(√3+1)]/(2) m
BC = 0.8(√3+1)
Thus, the height of the pedestal is 0.8(√3+1) m.
QUESTION 9
ANSWER
Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of
the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building
respectively.
In right ΔBCD,
tan 60° = CD/BC
√3 = 50/BC
BC = 50/√3 …(1)
Again,
In right ΔABC,
tan 30° = AB/BC
1/√3 = AB/BC
Use result obtained in equation (1)
AB = 50/3
Thus, the height of the building is 50/3 m.
QUESTION 10
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
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken. BD is the distance between the
poles.
From figure, AB = CD,
OB + OD = 80 m
In right ΔCDO,
tan 30° = CD/OD
1/√3 = CD/OD
CD = OD/√3 … (1)
In right ΔABO,
tan 60° = AB/OB
√3 = AB/(80-OD)
AB = √3(80-OD)
AB = CD (Given)
√3(80-OD) = OD/√3 (Using equation (1))
3(80-OD) = OD
240 – 3 OD = OD
4 OD = 240
OD = 60
Putting the value of OD in equation (1)
CD = OD/√3
CD = 60/√3
CD = 20√3 m
Also,
OB + OD = 80 m
⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m
and60 m respectively.
QUESTION 11
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.
ANSWER
Given, AB is the height of the tower.
DC = 20 m (given)
As per given diagram,
In right ΔABD,
tan 30° = AB/BD
1/√3 = AB/(20+BC)
AB = (20+BC)/√3 … (i)
Again,
In right ΔABC,
tan 60° = AB/BC
√3 = AB/BC
AB = √3 BC … (ii)
From equation (i) and (ii)
√3 BC = (20+BC)/√3
3 BC = 20 + BC
2 BC = 20
BC = 10
Putting the value of BC in equation (ii)
AB = 10√3
This implies, the height of the tower is 10√3 m and the width of the canal is 10 m.
QUESTION 12
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 AB be the building of height 7 m and EC be the height of the tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.
EC = DE + CD
Also, CD = AB = 7 m. and BC = AD
To Find: EC = Height of the tower
Design a figure based on given instructions:
In right ΔAB
tan 45° = AB/BC
1= 7/BC
BC = 7
Since BC = AD
So AD = 7
Again, from right triangle ADE,
tan 60° = DE/AD
√3 = DE/7
⇒ DE = 7√3 m
Now: EC = DE + CD
= (7√3 + 7)
= 7(√3+1)
Therefore, height of the tower is 7(√3+1) m.
QUESTION 13
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
Let AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
Draw a figure based on given instructions:
To Find: CD = distance between two ships
Step 1: From right triangle ABC,
tan 45° = AB/BC
1= 75/BC
BC = 75 m
Step 2: Form right triangle ABD,
tan 30° = AB/BD
1/√3 = 75/BD
BD = 75√3
Step 3: To find measure of CD, use results obtained in step 1 and step 2.
CD = BD – BC
= (75√3 – 75)
= 75(√3-1)
The distance between the two ships is 75(√3-1) m.
QUESTION 13
14. 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.
ANSWER
Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m.
To Find: Distance travelled by the balloon = DE = CE – CD
Let us redesign the given figure as per our convenient
Step 1: In right ΔBEC,
tan 30° = BE/CE
1/√3= 87/CE
CE = 87√3
Step 2:
In right ΔADC,
tan 60° = AD/CD
√3= 87/CD
CD = 87/√3 = 29√3
Step 3:
DE = CE – CD
= (87√3 – 29√3)
= 29√3(3 – 1) = 58√3
Distance travelled by the balloon = 58√3 m.
QUESTION 15
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 AB be the tower.
D is the initial and C is the final position of the car respectively.
Since the man is standing at the top of the tower so, Angles of depression are measured
from A.
BC is the distance from the foot of the tower to the car.
Step 1: In right ΔABC,
tan 60° = AB/BC
√3 = AB/BC
BC = AB/√3
AB = √3 BC
Step 2:
In right ΔABD,
tan 30° = AB/BD
1/√3 = AB/BD
AB = BD/√3
Step 3:
Form step 1 and Step 2, we have
√3 BC = BD/√3 (Since LHS are same, so RHS are also same)
3 BC = BD
3 BC = BC + CD
2BC = CD
or BC = CD/2
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6/2 = 3 sec.
QUESTION 16
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
Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As
per question,
In right ΔABC,
tan x = AB/BC
tan x = AB/4
AB = 4 tan x … (i)
Again, from right ΔABD,
tan (90°-x) = AB/BD
cot x = AB/9
AB = 9 cot x … (ii)
Multiplying equation (i) and (ii)
AB2 = 9 cot x × 4 tan x
AB2 = 36 (because cot x = 1/tan x
AB = ± 6
Since height cannot be negative, therefore, the height of the tower is 6 m.
Hence Proved.
Introduction
Trigonometry, a branch of mathematics that deals with the relationships between angles and sides of triangles, has numerous practical applications in various fields. In the Central Board of Secondary Education (CBSE) Class 10 curriculum, trigonometry is introduced as an important topic that enables students to solve real-world problems. This article aims to highlight some key applications of trigonometry in Class 10 CBSE and how it is utilized in different fields.Heights and Distances
Trigonometry is extensively used to determine heights and distances in real-life scenarios. By applying trigonometric ratios such as sine, cosine, and tangent, students can calculate the height of a building or a tree, the distance between two objects, or the width of a river or a lake. These applications find relevance in fields such as surveying, architecture, navigation, and astronomy.Angles of Elevation and Depression
Trigonometry is employed to measure angles of elevation and depression. For instance, in surveying, trigonometric concepts are used to determine the angle of elevation from the ground to the top of a hill or the angle of depression from a mountaintop to a specific point on the ground. These measurements are crucial for various applications, including constructing roads, bridges, and buildings.
Trigonometry plays a vital role in navigation and astronomy. Sailors and pilots rely on trigonometric calculations to determine their position, course, and distances during voyages or flights. Astronomers use trigonometry to calculate the distance between celestial objects, the size of planets, and the motion of stars.
Trigonometry is widely used in electrical engineering and physics. In electrical circuits, trigonometric principles help determine the magnitude and phase relationships of alternating current (AC) signals. In physics, trigonometry is applied to study the motion of objects, such as projectile motion, simple harmonic motion, and waves.
Trigonometry finds practical application in architecture and construction. Architects use trigonometric concepts to design buildings with precise angles and dimensions. Trigonometry enables accurate calculations for roof slopes, staircases, and other structural elements. Civil engineers employ trigonometry to calculate forces and stresses on bridges, buildings, and other structures.
In multimedia and animation, trigonometry plays a crucial role in creating realistic graphics and animations. Trigonometric functions such as sine and cosine help generate smooth and accurate motion, rotation, and scaling effects. Trigonometry is the backbone of computer graphics, allowing objects and characters to move and transform convincingly.
Trigonometry is utilized in sound and music applications. By analyzing the waveforms of sound waves, trigonometric concepts are employed to determine the frequency, amplitude, and phase of the wave. Trigonometry is also used in the design and construction of musical instruments, helping to create harmonious sounds and tones.
Conclusion
Trigonometry holds immense practical significance in various fields of study and real-world applications. In CBSE Class 10, students are introduced to the fundamental concepts of trigonometry, including trigonometric ratios, angles of elevation and depression, and solving problems involving heights and distances. Understanding these concepts equips students with the necessary skills to apply trigonometry to solve problems and analyze real-life scenarios in fields such as surveying, navigation, physics, engineering, animation, and architecture. By appreciating the relevance of trigonometry in diverse areas, students can develop a deeper understanding of the subject and recognize its practical importance beyond the classroom.
solved with the help of trigonometric ratios.
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