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# Class 10 NCERT Solutions- Chapter 9 Some Application of Trigonometry – Exercise 9.1 | Set 2

• Last Updated : 05 Apr, 2021

### Question 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 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.). Find the height of the tower and the width of the canal.

Solution:

In fig: AB = tower = ?

CB = canal = ?

In rt. âˆ†ABC,

tan60Â° =

h/x = âˆš3

h = âˆš3 x          -(1)

In rt. âˆ†ABD

= tan 30Â°

= 1/âˆš3

h = (x + 20)/âˆš3          -(2)

From (1) and (2)

âˆš3/1 = (x + 20)/âˆš3

3x = x + 20

3x – x = 20

2x = 20

X = 20/2

X = 10

Width of the canal is 10m

Putting value of x in equation 1

h = âˆš3 x

= 1.732(10)

= 17.32

Height of the tower 17.32m.

### Question 12. 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.

Solution:

In fig: ED = building = 7m

AC = cable tower = ?

In rt âˆ†EDC,

= tan45Â°

7/x = 1/1

DC = 7

Now, EB = DC = 7m

In rt. âˆ†ABE,

= tan60Â°

AB/7 = âˆš3/1

Height of tower = AC = AB + BC

7âˆš3 + 7

= 7(âˆš3 + 1)

= 7(1.732 + 1)

= 7(2.732)

Height of cable tower = 19.125m

### Question 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.

Solution:

In fig:

AB = lighthouse = 75m

D and C are two ships

DC = ?

In rt. âˆ†ABD,

= tan30Â°

75/BD = 1/âˆš3

BD = 75âˆš3

In rt. âˆ†ABC

= tan45Â°

75/BC = 1/1

BC = 75

DC = BD – BC

= 75âˆš3 – 75

75(âˆš3 – 1)

75(1.372 – 1)

34.900

Hence, distance between two sheep is 34.900

### Question 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 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.). Find the distance traveled by the balloon during the interval.

Solution:

In fig: AB = AC – BC

= 88.2 – 1.2

= 81m

In rt. âˆ†ABE

= 87/EB = tan30Â°

87/EB = 1/âˆš3

EB = 87âˆš3

In rt. âˆ†FDE

= tan60Â°

âˆš3 ED = 87

ED = 87/âˆš3

DB = DB – ED

87âˆš3 – 87/âˆš3

87(âˆš3 – 1/âˆš3)

= 87(3 – 1/âˆš3)

= 87(2/âˆš3) = 174/âˆš3 * âˆš3/âˆš3

= 174 * âˆš3/3 = 58âˆš3

58 * 1.732 = 100.456m

Distance traveled by balloon is 100.456m

### Question 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 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.

Solution:

In fig: AB is tower

In rt. âˆ†ABD

= tan30Â°

= 1/âˆš3

DB = âˆš3 AB          -(1)

In rt. âˆ†ABC

= tan60Â°

BC = AB/âˆš3          -(2)

DC = DB – BC

= âˆš3 AB – AB/âˆš3

AB(3 – 1/âˆš3)

CD = 2AB/âˆš3

S1 = S2

\frac{D1}{T1} = \frac{D2}{T2}

2/âˆš3AB/6 = AB/âˆš3/t

2t = 6

t = 6/2

t = 3sec

### Question 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.

Solution:

In fig: AB is tower

To prove: AB = 6m

Given: BC = 4m          DB = 9m

In âˆ†ABC

= tanÎ¸

AB/4 = tanÎ¸         -(1)

In âˆ†ABD

= tan (90Â°-Î¸)

AB/9 = 1/ tanÎ¸

9/AB = tanÎ¸         -(2)

From (1) and (2)

AB/4 = 9/AB

AB2 = 36

AB = âˆš36

AB = âˆš(6 * 6)

AB = 6m

Height of the tower is 6m.

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