NCERT Solutions for Class 10 Maths Ex 9.1

NCERT Solutions for Class 10 Maths Ex 9.1

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

Height of the tree = AB + AC

= 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

hx√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, hx√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:
In right-angled ∆ABC, we have

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.