## 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). **

**Solution: **In right-angled ∆ABC, we have

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*

√3*h* = 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* = 3*x*

4*x* = 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.**

**Solution:**

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.