# NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

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**Chapter 9 Some Application of Trigonometry Class 10 Maths NCERT Solutions**that will be useful in knowing the important points about the chapter. These NCERT Solutions for Class 10 Maths are updated according to the latest pattern of Syllabus. By regular practise of NCERT Solutions for Class 10, you will be able to solve the difficult problems in a given exercise.
Exercise: 9.1

**Page No: 203**

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 30Â° (see Fig. 9.11).Â

**Answer**

Let AB be the vertical pole Ac be 20 m Â long rope tied to point C.

InÂ Â right Î”ABC,

sin 30Â° = AB/AC

â‡’ 1/2 = AB/20

â‡’ AB = 20/2

â‡’ AB = 10

The height of the pole is 10 m.

2.Â A tree breaks due to storm and the broken partÂ bends so that the top of the tree touches the groundÂ making an angle 30Â° with it. The distance betweenÂ the foot of the tree to the point where the topÂ touches the ground is 8 m. Find the height of theÂ tree.

**Answer**

InÂ Â right Î”ABC,

cos 30Â° = BC/AC

â‡’Â âˆš3/2 = 8/AC

â‡’ AC = 16/âˆš3

Also,

tan 30Â° = AB/BC

â‡’ 1/âˆš3Â = AB/8

â‡’ AB = 8/âˆš3

Total height of the tree = AB+AC = 16/âˆš3Â + 8/âˆš3 = 24/âˆš3

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 30Â° 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 60Â° to the ground. What should be the length of the slide in each case?

**Answer**

There are two slides of height 1.5 m and 3 m. (Given)

Let AB is 1.5 m and PQ be 3 m slides.Â

ABC is the slide inclined at 30Â° with length AC and PQR is the slide inclined atÂ

60Â° with length PR.

A/q,

InÂ Â right Î”ABC,

sin 30Â° = AB/AC

â‡’ 1/2 = 1.5/AC

â‡’ AC = 3m

also,

InÂ Â right Î”PQR,

sin 60Â° = PQ/PR

â‡’Â âˆš3/2 = 3/PR

â‡’ PR = 2âˆš3 m

Hence, length of the slides are 3 m and 2âˆš3 m respectively.

**Page No: 204**

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 30Â°. 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.

A/q,

InÂ Â right Î”ABC,

tan 30Â° = AB/BC

â‡’ 1/âˆš3Â = AB/30

â‡’ 1/âˆš3Â = AB/30

â‡’ AB = 10âˆš3

Thus, the height of the tower is 10âˆš3 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 60Â°. Find the length of the string, assuming that there is no slack in the string.

Let BC be the height of the kite from the ground,

AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.

Thus, the height of the tower is 10âˆš3 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 60Â°. Find the length of the string, assuming that there is no slack in the string.

**Answer**Let BC be the height of the kite from the ground,

AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.

A/q,

InÂ Â right Î”ABC,

sin 60Â° = BC/AC

â‡’Â âˆš3/2 = 60/AC

â‡’ AC = 40âˆš3 m

Thus, the length of the string from the ground is 40âˆš3 m.

6. Â 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 30Â° to 60Â° as he walks towards the building. Find the distance he walked towards the building.

Let the boy initially standing at point Y with inclination 30Â° and then he approaches the building to

the point X with inclination 60Â°.

âˆ´ XY is the distance he walked towards the building.

also, XY = CD.

Height of the building = AZ = 30 m

Thus, the length of the string from the ground is 40âˆš3 m.

6. Â 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 30Â° to 60Â° as he walks towards the building. Find the distance he walked towards the building.

**Answer**Let the boy initially standing at point Y with inclination 30Â° and then he approaches the building to

the point X with inclination 60Â°.

âˆ´ XY is the distance he walked towards the building.

also, XY = CD.

Height of the building = AZ = 30 m

AB = AZ - BZ = (30 - 1.5) = 28.5 m

A/q,

A/q,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3Â = 28.5/BD

â‡’ 1/âˆš3Â = 28.5/BD

â‡’ BD = 28.5âˆš3 m

also,

InÂ Â right Î”ABC,

also,

InÂ Â right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3Â = 28.5/BC

â‡’ âˆš3Â = 28.5/BC

â‡’ BC = 28.5/âˆš3 = 28.5âˆš3/3 m

âˆ´ XY = CD = BD - BC = (28.5âˆš3 - 28.5âˆš3/3) = 28.5âˆš3(1-1/3) = 28.5âˆš3Â Ã— 2/3 = 57/âˆš3 m.

Thus, the distance boy walked towards the building isÂ 57/âˆš3 m.

7. Â 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 45Â° and 60Â° respectively. Find the height of the tower.

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

A/q,

âˆ´ XY = CD = BD - BC = (28.5âˆš3 - 28.5âˆš3/3) = 28.5âˆš3(1-1/3) = 28.5âˆš3Â Ã— 2/3 = 57/âˆš3 m.

Thus, the distance boy walked towards the building isÂ 57/âˆš3 m.

7. Â 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 45Â° and 60Â° 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

A/q,

InÂ Â right Î”BCD,

tan 45Â° = BC/CD

â‡’ 1Â = 20/CD

â‡’ 1Â = 20/CD

â‡’ CD = 20 m

also,

also,

InÂ Â right Î”ACD,

tan 60Â° = AC/CD

â‡’Â âˆš3Â = AC/20

â‡’Â âˆš3Â = AC/20

â‡’ AC = 20âˆš3 m

Height of transmission tower = AB = AC - BC = (20âˆš3 - 20) m = 20(âˆš3 - 1) m.

8. Â 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 60Â° and from the same point the angle of elevation of the top of the pedestal is 45Â°. Find the height of the pedestal.

Let AB be the height of statue.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC - BC = (20âˆš3 - 20) m = 20(âˆš3 - 1) m.

8. Â 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 60Â° and from the same point the angle of elevation of the top of the pedestal is 45Â°. 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.

Height of pedestal = BC = AC - AB

A/q,

InÂ Â right Î”BCD,

tan 45Â° = BC/CD

â‡’Â 1Â = Â BC/CD

tan 45Â° = BC/CD

â‡’Â 1Â = Â BC/CD

â‡’ BC = CD.

also,

also,

InÂ Â right Î”ACD,

tan 60Â° = AC/CD

â‡’Â âˆš3Â = AB+BC/CD

â‡’Â âˆš3Â = AB+BC/CD

â‡’ âˆš3CD = 1.6 mÂ + BC

â‡’ âˆš3BC = 1.6 mÂ + BC

â‡’ âˆš3BC - BC = 1.6 m

â‡’ BC(âˆš3-1) = 1.6 m

â‡’ BC = 1.6/(âˆš3-1) m

â‡’ BC = 0.8(âˆš3+1)Â m

Thus,Â the height of the pedestal is 0.8(âˆš3+1)Â m.

9. The angle of elevation of the top of a building from the foot of the tower is 30Â° and the angle of elevation of the top of the tower from the foot of the building is 60Â°. If the tower is 50 m high, find the height of the building.

Let CD be the height of the tower equal to 50 m (Given)

Let AB be the height of the building.

BC be the distance between the foots of the building and the tower.

Elevation is 30Â° and 60Â° from the tower and the building respectively.

A/q,

â‡’ âˆš3BC = 1.6 mÂ + BC

â‡’ âˆš3BC - BC = 1.6 m

â‡’ BC(âˆš3-1) = 1.6 m

â‡’ BC = 1.6/(âˆš3-1) m

â‡’ BC = 0.8(âˆš3+1)Â m

Thus,Â the height of the pedestal is 0.8(âˆš3+1)Â m.

9. The angle of elevation of the top of a building from the foot of the tower is 30Â° and the angle of elevation of the top of the tower from the foot of the building is 60Â°. If the tower is 50 m high, find the height of the building.

**Answer**Let CD be the height of the tower equal to 50 m (Given)

Let AB be the height of the building.

BC be the distance between the foots of the building and the tower.

Elevation is 30Â° and 60Â° from the tower and the building respectively.

A/q,

InÂ Â right Î”BCD,

tan 60Â° = CD/BC

â‡’Â âˆš3Â = 50/BC

â‡’Â âˆš3Â = 50/BC

â‡’ BC = 50/âˆš3

also,

also,

InÂ Â right Î”ABC,

tan 30Â° = AB/BC

â‡’ 1/âˆš3Â = AB/BC

â‡’ 1/âˆš3Â = AB/BC

â‡’ AB = 50/3

Thus, the height of the building is 50/3.

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 60Â° and 30Â°, respectively. Find the height of the poles and the distances of the point from the poles.

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.

A/q,

AB = CD,

OBÂ + OD = 80 m

Now,

Thus, the height of the building is 50/3.

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 60Â° and 30Â°, 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.

A/q,

AB = CD,

OBÂ + OD = 80 m

Now,

InÂ Â right Î”CDO,

tan 30Â° = CD/OD

â‡’ 1/âˆš3Â = CD/OD

â‡’ 1/âˆš3Â = CD/OD

â‡’Â CDÂ = OD/âˆš3 ... (i)

also,

also,

InÂ Â right Î”ABO,

tan 60Â° = AB/OB

â‡’Â âˆš3Â = AB/(80-OD)

â‡’Â âˆš3Â = AB/(80-OD)

â‡’ AB = âˆš3(80-OD)

AB = CD (Given)

â‡’ âˆš3(80-OD) = OD/âˆš3

â‡’ 3(80-OD) = OD

â‡’ 240 - 3 OD = OD

â‡’ 4 OD = 240

â‡’ OD = 60

Putting the value of OD in equation (i)

CD = OD/âˆš3 â‡’ CD = 60/âˆš3 â‡’ CD = 20âˆš3 m

also,

OBÂ + OD = 80 m â‡’ OB = (80-60) m = 20 m

Thus, the height of the poles are 20âˆš3 m and distance from the point of elevation are 20 m and 60 m respectively.

11. Â 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 60Â°. 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 30Â° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Here, AB is the height of the tower.

CD = 20 m (given)

A/q,

AB = CD (Given)

â‡’ âˆš3(80-OD) = OD/âˆš3

â‡’ 3(80-OD) = OD

â‡’ 240 - 3 OD = OD

â‡’ 4 OD = 240

â‡’ OD = 60

Putting the value of OD in equation (i)

CD = OD/âˆš3 â‡’ CD = 60/âˆš3 â‡’ CD = 20âˆš3 m

also,

OBÂ + OD = 80 m â‡’ OB = (80-60) m = 20 m

Thus, the height of the poles are 20âˆš3 m and distance from the point of elevation are 20 m and 60 m respectively.

11. Â 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 60Â°. 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 30Â° (see Fig. 9.12). Find the height of the tower and the width of the canal.

**Answer**Here, AB is the height of the tower.

CD = 20 m (given)

A/q,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3Â = AB/(20+BC)

â‡’ 1/âˆš3Â = AB/(20+BC)

â‡’ AB = (20+BC)/âˆš3 ... (i)

also,

InÂ Â right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3Â = AB/BC

also,

InÂ Â right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3Â = AB/BC

â‡’ AB = âˆš3 BC ... (ii)

From eqn (i) and (ii)

AB = âˆš3 BC = (20+BC)/âˆš3

â‡’ 3 BC = 20Â + BC

â‡’ 2 BC = 20 â‡’ BC = 10 m

Putting the value of BC in eqn (ii)

AB = 10âˆš3 m

Thus, the height of the tower 10âˆš3 m and the width of the canal is 10 m.

12. Â From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60Â° and the angle of depression of its foot is 45Â°. Determine the height of the tower.

Let AB be the building of height 7 m and EC be the height of tower.

A is the point from where elevation of tower is 60Â° and the angle of depression of its foot is 45Â°

EC = DEÂ + CD

also, CD = AB = 7 m.

and BC = AD

A/q,

InÂ Â right Î”ABC,

tan 45Â° = AB/BC

â‡’ 1= 7/BC

â‡’Â BC = 7 m = AD

also,

InÂ Â right Î”ADE,

tan 60Â° = DE/AD

â‡’Â âˆš3Â = DE/7

â‡’ DE = 7âˆš3 m

Height of the tower = EC = Â DEÂ + CD

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (7âˆš3Â + 7) m = 7(âˆš3+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 30Â° and 45Â°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Let AB be the lighthouse of height 75 m.

Let C and D be the positions of the ships.

30Â° and 45Â° are the angles of depression from the lighthouse.

A/q,

InÂ Â right Î”ABC,

tan 45Â° = AB/BC

â‡’ 1= 75/BC

â‡’Â BC = 75 m

also,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3 = 75/BD

â‡’Â BD = 75âˆš3Â Â m

The distance between the two ships = CD = BD - BC = (75âˆš3 - 75) m = 75(âˆš3-1) m.

14. Â 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 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 60Â°. After some time, the angle of elevation reduces to 30Â° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

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

Distance travelled by the balloon =

Â DE = CE - CD

A/q,

InÂ Â right Î”BEC,

tan 30Â° = BE/CE

â‡’ 1/âˆš3= 87/CE

â‡’ CE = 87âˆš3Â m

also,

InÂ Â right Î”ADC,

tan 60Â° = AD/CD

â‡’ âˆš3= 87/CD

â‡’ CD = 87/âˆš3Â m = 29âˆš3Â m

Distance travelled by the balloon = Â DE = CE - CD = (87âˆš3 - 29âˆš3) m = 29âˆš3(3 - 1) m = 58âˆš3 m.

15. Â 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 30Â°, 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 60Â°. Find the time taken by the car to reach the foot of the tower from this point.

Let AB be the tower.

D is the initial and C is the final position of the car respectively.

Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

A/q,

InÂ Â right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3 = AB/BC

â‡’Â BCÂ = AB/âˆš3Â m

also,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3 = AB/(BC + CD)

â‡’ ABâˆš3 = BC + CD

â‡’ ABâˆš3 =Â AB/âˆš3Â + CD

â‡’ CD = ABâˆš3 - AB/âˆš3

â‡’ CD = AB(âˆš3 - 1/âˆš3)

â‡’ CD = 2AB/âˆš3

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.

16. Â 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 AB be the tower.

C and D be the two points with distance 4 m and 9 m from the base respectively.

A/q,

InÂ Â right Î”ABC,

tan

â‡’Â tanÂ

â‡’Â ABÂ = 4 tanÂ

also,

InÂ Â right Î”ABD,

tanÂ (90Â°-

â‡’ cotÂ

â‡’Â ABÂ =Â 9Â cotÂ

Multiplying Â eqn (i) and (ii)

AB

â‡’ AB

â‡’ AB =Â Â± 6

Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

From eqn (i) and (ii)

AB = âˆš3 BC = (20+BC)/âˆš3

â‡’ 3 BC = 20Â + BC

â‡’ 2 BC = 20 â‡’ BC = 10 m

Putting the value of BC in eqn (ii)

AB = 10âˆš3 m

Thus, the height of the tower 10âˆš3 m and the width of the canal is 10 m.

12. Â From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60Â° and the angle of depression of its foot is 45Â°. Determine the height of the tower.

**Answer**Let AB be the building of height 7 m and EC be the height of tower.

A is the point from where elevation of tower is 60Â° and the angle of depression of its foot is 45Â°

EC = DEÂ + CD

also, CD = AB = 7 m.

and BC = AD

A/q,

InÂ Â right Î”ABC,

tan 45Â° = AB/BC

â‡’ 1= 7/BC

â‡’Â BC = 7 m = AD

also,

InÂ Â right Î”ADE,

tan 60Â° = DE/AD

â‡’Â âˆš3Â = DE/7

â‡’ DE = 7âˆš3 m

Height of the tower = EC = Â DEÂ + CD

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (7âˆš3Â + 7) m = 7(âˆš3+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 30Â° and 45Â°. 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.

30Â° and 45Â° are the angles of depression from the lighthouse.

A/q,

InÂ Â right Î”ABC,

tan 45Â° = AB/BC

â‡’ 1= 75/BC

â‡’Â BC = 75 m

also,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3 = 75/BD

â‡’Â BD = 75âˆš3Â Â m

**Page No: 205**14. Â 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 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 60Â°. After some time, the angle of elevation reduces to 30Â° (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

Distance travelled by the balloon =

Â DE = CE - CD

A/q,

InÂ Â right Î”BEC,

tan 30Â° = BE/CE

â‡’ 1/âˆš3= 87/CE

â‡’ CE = 87âˆš3Â m

also,

InÂ Â right Î”ADC,

tan 60Â° = AD/CD

â‡’ âˆš3= 87/CD

â‡’ CD = 87/âˆš3Â m = 29âˆš3Â m

Distance travelled by the balloon = Â DE = CE - CD = (87âˆš3 - 29âˆš3) m = 29âˆš3(3 - 1) m = 58âˆš3 m.

15. Â 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 30Â°, 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 60Â°. 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.

Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

A/q,

InÂ Â right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3 = AB/BC

â‡’Â BCÂ = AB/âˆš3Â m

also,

InÂ Â right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3 = AB/(BC + CD)

â‡’ ABâˆš3 = BC + CD

â‡’ ABâˆš3 =Â AB/âˆš3Â + CD

â‡’ CD = ABâˆš3 - AB/âˆš3

â‡’ CD = AB(âˆš3 - 1/âˆš3)

â‡’ CD = 2AB/âˆš3

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.

16. Â 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.

A/q,

InÂ Â right Î”ABC,

tan

*x*= AB/BCâ‡’Â tanÂ

*xÂ*= AB/4â‡’Â ABÂ = 4 tanÂ

*x*... (i)also,

InÂ Â right Î”ABD,

tanÂ (90Â°-

*x*)Â = AB/BDâ‡’ cotÂ

*xÂ*= AB/9â‡’Â ABÂ =Â 9Â cotÂ

*xÂ*Â ... (ii)Multiplying Â eqn (i) and (ii)

AB

^{2}= 9Â cotÂ*xÂ*Ã— 4 tanÂ*x*â‡’ AB

^{2}Â = 36â‡’ AB =Â Â± 6

Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

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### NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

These Chapter 9 Some Applications of Trigonometry Class 10 Maths NCERT Solutions that will be helpful in Grasping the key concepts inside the chapter properly. There are only 2 topics and only one exercise which will be enhance your knowledge about the applications of trigonometry in real life.

â€¢ Introduction: Trigonometry is used in our daily lives. It is used in astronomy, geography and in navigation. The knowledge of trigonometry is used to construct maps, determine the position of an island in relation to the longitudes and latitudes.

â€¢Â Heights and Distances: In this topic, we will understand about various terms such asÂ line of sight, angle of elevation and angle of depression and discuss various problems.

You can also access other study resources prepared by Studyrankers experts:

You can also find exercisewise NCERT Solutions for Chapter 9 Some Applications of Trigonometry by clicking on the given below.

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**FAQ onÂ Chapter 9 Some Application of Trigonometry**

#### How many exercises in Chapter 9 Some Applications of Trigonometry

There are only 1 exercise in the Chapter 9 Class 10 Maths that will encourage students to learn topics.

#### What isÂ Line of sight?

A line drawn from the eye of the observer to the point in the object viewed by the observer is called the line of sight.

#### What is Angle of elevation?

he angle formed by a line of sight OP with the horizontal line OX when the object is above the horizontal line is called the angle of elevation.

#### What isÂ Angle of depression?

The angle formed by a line of sight OP with the horizontal line OX when the object is below the horizontal line is called angle of depression.