In this post, you will find the NCERT solutions for class 10 maths ex 9.1. These solutions are based on the latest curriculum of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 9.1
1. 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 figure).
sin 30° = AB/AC
1/2 = AB/20
AB = 10 m
Hence, the height of the pole is 10 m.
2. 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.
Solution:
In right-angled ∆ABC, we have
cos 30° = BC/AC
√3/2 = 8/AC
AC = 16/√3 m
Again, tan 30° = AB/BC
1/√3 = AB/8
AB
= 8/√3 m
= 8/√3 + 16/√3 = 24/√3
= 24/√3 × √3/√3
= 8√3 m
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
of 3 m and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case?
Solution:
In right-angled ∆ABC, we have
sin 30° = AB/AC
1/2 = 1.5/AC
AC
= 3 m
In
right-angled ∆PQR, we have
sin 60° = PQ/PR
√3/2= 3/PR
PR
= 2√3 m
Hence,
the lengths of the slides are 3 m and 2√3 m
respectively.
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.
Solution:
In right-angled ∆ABC, we have
tan 30° = AB/BC
1/√3 = AB/30
AB
= 30/√3 m
= 30/√3 ×
√3/√3 = 10√3 m
Hence,
the height of the tower is 10√3 m.
5. 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.
Solution:
In right-angled ∆ABC, we have
sin 60° = AB/AC
√3/2 = 60/AC
AC
= 40√3 m
Hence, the length of the string is 40√3 m.
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.
Solution: Here,
the height of the building AB
= 30 m and the height of the boy PR = 1.5 m
AC
= AB – BC
= AB – PR
= 30 – 1.5
= 28.5 m
In
right-angled ∆ACQ, we have
tan
60° = AC/QC
√3
= 28.5/QC
QC
= 28.5/√3 m
In
right-angled ∆ACP, we have
tan
30° = AC/PC
1/√3 =
28.5/PC
PC
= 28.5√3 m
Now,
PQ = PC – QC
= 28.5√3
– 28.5/√3
= (85.5 – 28.5)/√3
= 57/√3
= 57/√3 × √3/√3
PQ
= 19√3 m
Hence,
the boy walked towards the building is 19√3 m.
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.
Solution: In the given figure,
AC is the tower and AB is the building.
Let the height of the tower be h metres.
In right-angled ∆CBP, we have
tan
60° = BC/BP
√3
= (AB + AC)/BP
√3 =
(20 + h)/BP …………. (i)
In
right-angled ∆ABP, we have
tan
45° = AB/BP
1 = 20/BP
BP
= 20 m
Putting
the value of BP in equation (i), we get
√3 = (20
+ h)/20
20√3 =
20 + h
h = 20√3 – 20
h = 20(√3 – 1) m
Hence, the
height of the tower is 20(√3 – 1) m.
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.
Solution: In
the given figure, AB is the statue and BC is the pedestal.
Let the height of the pedestal be h metres.
Therefore, BC
= h metres
In
right-angled ∆ACP, we have
tan
60° = AC/PC
√3
= (AB + BC)/PC
√3 =
(1.6 + h)/PC ………. (i)
In
right-angled ∆BCP, we have
tan
45° = BC/PC
1
= h/PC
So, PC
= h ……….. (ii)
From equations (i) and (ii), we get
√3 =
(1.6 + h)/h
√3h = 1.6 + h
h(√3 – 1) = 1.6
h = 1.6/(√3 –
1)
h = 1.6/(√3 –
1) × (√3 + 1)/(√3 + 1)
h = 1.6(√3 +
1)/(3 – 1)
h = 1.6(√3 +
1)/2
h = 0.8(√3 +
1) m
Hence,
the height of the pedestal is 0.8(√3 +
1) m.
9. 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.
Solution: Let the height of the building
be h metres.
In
right-angled ∆PQB, we have
tan
60° = PQ/BQ
√3 =
50/BQ
BQ
= 50/√3 m ………. (i)
In
right-angled ∆ABQ, we have
tan
30° = AB/BQ
1/√3 = h/BQ
BQ
= h√3 m
………. (ii)
From
equations (i) and (ii), we have
h√3 = 50/√3
h = 50/3
h = 16 ⅔ m
Hence,
the height of the building is 16
⅔ m.
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.
Solution:
In right-angled ∆PRQ, we have
tan
60° = PQ/QR
√3 = h/x
h = x√3 m ………. (i)
In
right-angled ∆ABR, we have
tan
30° = AB/BR
1/√3
= h/(80 – x)
1/√3
= x√3/(80 – x) [From equation (i)]
80
– x = 3x
4x = 80
x = 20 m
Therefore, h = x√3 = 20√3 m
Also,
BR = 80 – x = 80 – 20 = 60
m
Hence, the
heights of the poles are 20√3 m
each and the distances of the point from poles are 20 m and 60 m, respectively.
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 figure). Find the
height of the tower and the width of the canal.
Solution: In right-angled ∆ABC, we have
tan
60° = AB/BC
√3
= AB/BC
AB
= BC√3 m ………. (i)
In
right-angled ∆ABD, we have
tan
30° = AB/BD
1/√3
= AB/(BC + CD)
1/√3
= AB/(BC + 20)
AB
= (BC + 20)/√3 m ………. (ii)
From
equations (i) and (ii), we get
BC√3 = (BC + 20)/√3
3BC
= BC + 20
BC
= 10 m
From
equation (i), AB = 10√3 m
Hence,
the height of the tower is 10√3 m
and the width of the canal is 10 m.
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.
Solution:
In right-angled ∆ABD, we have
tan
45° = AB/BD
1
= 7/BD
BD
= 7 m
AE
= 7 m
In
right-angled ∆AEC, we have
tan
60° = CE/AE
√3
= CE/7
CE
= 7√3 m
CD
= CE + ED
= CE + AB
= 7√3 +
7 = 7(√3 + 1) m
Hence,
the height of the tower is 7(√3 + 1) m.
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
two ships.
Solution:
In right-angled ∆ABQ, we have
tan
45° = AB/BQ
1
= 75/BQ
BQ
= 75 m ………. (i)
In
right-angled ∆ABP, we have
tan
30° = AB/BP
1/√3 = AB/(BQ + QP)
1/√3 = 75/(75 + QP) [From equation (i)]
75
+ QP = 75√3
QP
= 75(√3 – 1) m
Hence,
the distance between the two ships is 75(√3
– 1) m.
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 figure). Find the
distance travelled by the balloon during the interval.
tan
60° = AB/BC
√3 = 88.2/BC
BC
= 88.2/√3 m
In
right-angled ∆PQC, we have
tan
30° = PQ/CQ
1/√3 = 88.2/CQ
CQ = 88.2√3
Now,
BQ = CQ – CB
= 88.2√3 – 88.2/√3
= (264.6
– 88.2)/√3
= 176.4/√3
BQ
= 176.4/√3 × √3/√3
= 58.8√3 m
Hence
the distance travelled by the balloon during the interval is 58.8√3 m.
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.
Solution:
In right-angled ∆ABP, we have
tan
30° = AB/BP
1/√3 = AB/BP
BP
= AB√3 ………. (i)
In
right-angled ∆ABQ, we have
tan
60° = AB/BQ
√3
= AB/BQ
BQ
= AB/√3 ……….
(ii)
Therefore, PQ
= BP – BQ
= AB√3 – AB/√3
= (3AB – AB)/√3
= 2AB/√3
= 2BQ
[From eq. (ii)]
BQ
= ½ PQ
Time
taken by the car to travel a distance PQ = 6 seconds
Time
taken by the car to travel a distance BQ, i.e., ½ PQ = ½ × 6 = 3 seconds
Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.