**Download NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1 – Some Applications of Trignometry**

**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°.**

Let the height of the pole AB be h.

In ΔABC, ∠B = 90°

sin 30° = =

=

h = 10 m

Hence, the height of the pole is 10m.

**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.**

*Let the tree***AC’**broke at point**A**and the broken part**AC’**bends over such that it touches the ground at**C**

AC’ = AC

In ΔABC, ∠B = 90°

tan 30° =

m

Similarly, cos 30° =

m

Height of the tree = BA + AC’ = BA + AC

= + m

=

= 8√3m

Hence, the height of the tree is 8√3m

**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 slide in each case?**

*Let a slide has its top at a height of***h**meters and is inclined at an angle**θ**to the ground. Let**l**denote the length of slide

In ΔABC, ∠B = 90°

sin θ = =

For children below age of 5 years, h = 1.5 m, θ = 30°

For elder children, h = 3 m, θ = 60°

= 2√3m

**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.**

*Consider the tower***AB**and the point**C**on the ground, 30 m away from the foot**B**of the tower. Let the height of the tower be**h**

In ΔABC, ∠B = 90°

tan 30° =

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.**

*Consider the kite flying at point***A**and the string held on the ground at point**C**. Let the length of the string be**l**

In ΔABC, ∠B = 90°

sin 60° =

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 on the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.**

*Consider the***AC**,**B**is the point in line of sight of the boy. Then, he moves from**PQ**to**RS**

∴ PQ = RS = BC = 1.5m

AB = AC – BC = 30 – 1.5 = 28.5m

In ΔAPB, ∠B = 90°

tan 30° =

BP = AB√3

In ΔABR, ∠B = 90°

tan 60° =

√3 =

BR =

Distance travelled by boy = BP – BR

= AB√3 –

= 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° Find the height of the tower.**

*Consider the tower***AB**mounted on the top of the building**BC**and the point**D**on the ground

In ΔBCD, ∠C = 90°

tan 45° =

1 =

∴ DC = 20m

In ΔACD, ∠C = 90°

tan 60° =

√3 =

AC = 20√3m

Height of the tower = AC – BC

= 20√3 – 20m

= 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.**

*Let***AB**be the statue, mounted on the pedestal**BC**of height**h**

In ΔBCD, ∠C = 90°

tan 45° =

1 =

∴ DC = h

In ΔACD, ∠C = 90°

tan 60° =

√3 =

Putting DC = h

√3 =

√3h = 1.6 + h

h=

h=

h=

h = 0.8(√3 + 1) m

**10. 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.**

*Let***AB**be a building of height**h**and**CD**be the tower of height 50 m

In ΔCDB, ∠D = 90°

tan 60° =

√3 =

DB=

In ΔADB, ∠B = 90°

tan 30° =

Putting, DB=

h =

**10. Two poles of equal height 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.**

*Consider poles to be***AB**and**CD**of height**h**. Point**P**is located on the road as shown

In ΔCDP, ∠D = 90°

tan 30° =

DP = h√3

In ΔABP, ∠B = 90°

tan 60° =

√3 =

Width of the road = DB = DP + BP

∴80 = h√3 +

80 =

80 =

h = 20√3 m

DP = h√3 = 60 m

BP = = 20m

*The height of the poles is 20√3*

*and the point*

**P**is at a distance of**60 m***from one pole and*

**20 m***from the other.*

**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 if 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°. Find the height of the tower and width of the canal.**

*Consider the tower***AB, C**is the point on the opposite bank of the canal and the point**D**be as indicated. Let the height of the tower be**h**

In ΔABD, ∠B = 90°

tan 30° =

DB = h√3

In ΔABC, ∠B = 90°

tan 60° =

√3 =

Distance between D and C = DB – CB

∴20 = h√3 –

20 =

20 =

h = 10√3 m

Width of the canal = CB = = 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.**

*Let the building be*∴PD = AB = 7m**AB**. The point**P**lies on the tower**CD**in the same horizontal line as the top**A**of the building

In ΔAPD, ∠P = 90°

tan 45° =

1 =

AP = 7m

In ΔACP, ∠P = 90°

tan 60° =

√3 =

CP = 7√3 m

Height of the tower = CP + PD = 7 + 7√3 = 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 the two ships.**

*Consider the lighthouse***AB**of height 75 m.**AX***be the horizontal.***D**and**C**are the positions of two ships.

∴AX // DB

∴∠ADB = ∠DAX = 30**° (Alternate angles)**

∠ACB = ∠CAX = 45**°**

In ΔABD, ∠B = 90°

tan 30° =

BD = 75√3 m

In ΔABC, ∠B = 90°

tan 45° =

1 =

BC = 75 m

Distance between the two ships = BD – BC

= 75√3 – 75

= 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°. Find the distance travelled by the balloon during the interval.**

*Let the girl be denoted by***PQ**.**A**and**C**are the initial and final positions of balloon, respectively.**PBD**is a horizontal line where**B**and**D**lie directly below**A**and**C**.**E**is a point on the ground below**C**PQ = DE = 1.2 m

AB = CD = CE – DE = 88.2 – 1.2 = 87 m

In ΔPCD, ∠D = 90°

tan 30° =

tan 30° =

PD = 87√3 m

In ΔABP, ∠B = 90°

tan 60° =

√3 = [AB = CD = 87m]

Distance travelled by the balloon = AC = BD = PD – PB

= 87√3 –

=

=

= 58√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.**

*Consider the tower***AB**of height**h**.**AX***be the horizontal.***D**is the initial position and**C**is the position of the car after six seconds.

∴AX // DB

∴∠ADB = ∠DAX = 30**° (Alternate angles)**

∠ACB = ∠CAX = 60**°**

In ΔABD, ∠B = 90°

tan 30° =

BD = h√3 m

In ΔABC, ∠B = 90°

tan 60° =

√3 =

Distance between D and C = DB – CB

= h√3 –

=

=

Speed of the car = = =

Time taken to travel remaining (BC = ) distance =

=

= 3 s

**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.**

*Let the height of the tower***AB**be**h**. Point**C**and**D**are*and**away from the foot of the tower. Let the angle of elevation of the top of the tower from***D**be**α**and from**C**be**β***and**are complementary***∴α**and**β***and**are complementary***α****+ β = 90°**

In ΔABD, ∠B= 90°

tan *α* = = ………(i)

In ΔABC, ∠B= 90°

tan *β* = =

tan (90° – *α)* = (**∴ ****α** **+ β = 90°)**

cot *α* = (**∴ ****tan (90° – θ)**** = cotθ)**

=

Putting value from equation (i)

=

h² = 36

h = ±6m

Height cannot be negative

Hence, the height of the tower is 6m.

**Download NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1 – Some Applications of Trignometry**

## Leave a Reply