Some Applications of Trigonometry – Exercise – Class 10

continued part of Some Applications of Trigonometry -Class 10

10.Two poles of equal heights are standing opposite each other and 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 30º, respectively. Find the height of poles and the distance of the point from the poles.

Solution:

Some Applications of Trigonometry – Exercise 9.2-Class 10

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ∆ABO,

AB/BO = tan 60˚

AB/BO = √3

BO = AB/√3

In ∆CDO,

CD/DO = tan 60˚

CD/80-BO = 1/√3

CD√3 = 80 – BO

CD√3 = 80 – AB/√3

CD√3 + AB/√3 = 80

Since the poles are of equal heights,

CD = AB

CD[√3 + 1/√3] = 80

CD[√3 + 1/√3] = 80

CD = 20 √3m

BO = AB/√3 = 20√3/√3 = 20 m

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles is 20√3 m and the point is 20 m and 60 m far from these poles.


  1. 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°. Find the height of the tower and the width of the canal.

Solution:

Some Applications of Trigonometry – Exercise 9.2-Class 10

In ∆ABC,

AB/BC = tan60˚

AB/BC = √3

BC = AB/√3

In ∆ABD,

AB/BD = tan30˚

AB/BC+CD = 1/√3

AB/[(AB/√3)+20] = AB/√3

AB√3/[(AB+20√3] = 1/√3

3AB = AB + 20√3

2AB = 20√3

AB = 10√3 m

BC = AB/√3 = (10√3/√3)m = 10m

Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.


  1. 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:

Some Applications of Trigonometry – Exercise 9.2-Class 10

Let AB be a building and CD be a cable tower.

In ∆ABD,

AB/BD = tan45˚

7/BD = 1

BD = 7M

In ∆ACE,

AC = BD = 7 m

CE/AE = tan60˚

CE/7 = √3

CE = 7√3

CD = CE + ED = 7√3m + 7m = 7(√3+1)m

Therefore, the height of the cable tower is 7(√3 + 1) m.


  1. 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.

Solution:

some-applications-of-trigonometry-exercise-9-14

Let AB be the lighthouse and the two ships be at point C and D respectively. In ∆ABC,

,

AB/BC = tan45˚

75/BC = 1

BC = 75 m

In ∆ABD

AB/BD = tan30˚

75/75+CD = 1/√3

75√3= 75 + CD

75(√3 – 1)m = CD

Therefore, the distance between the two ships is 75(√3 − 1) m


  1. 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 traveled by the balloon during the interval.

Solution:

some-applications-of-trigonometry-exercise-9-15

Let the initial position A of balloon change to B after some time and CD be the girl.

In ∆ACE,

AE/CE = tan60˚

AF-EF/CE = √3

88.2-1.2/CE = √3

CE = 87/√3 = 29√3m

In ∆BCG,

BG/CG = tan30˚

88.2-1.2/CG = 1/√3

87√3m = CG

Distance travelled by balloon = EG = CG – CE

= (87√3 – 29√3)m

= 58√3m


  1. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as 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:

Some Applications of Trigonometry – Exercise 9.2-Class 10

Let AB be the tower. Initial position of the car is C, which changes to D after six seconds.

In ∆ADB,

AB/DB = tan60˚

AB/DB = √3

DB = AB/√3

In ∆ABC,

AB/BC = tan30˚

AB/BC+CD = 1/√3

AB√3= BD + DC

AB√3= AB/√3 + DC

DC = AB√3 – AB/√3 = AB(√3-1/√3)

= 2AB/√3

Time taken by the car to travel distance DC = (= 2AB/√3) 6 seconds,

time taken by the car to travel distance DB (= 2AB/√3) = 6/(2AB/√3) X AB/√3 = 6/2 = 3 seconds


  1. 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:

Some Applications of Trigonometry – Exercise 9.2-Class 10

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively. The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ∆AQR,

AQ/QR = tan θ

AQ/4 = tanθ ———–(i)

In ∆AQS,

AQ/SQ = tan(90- θ)

AQ/9 = cot θ ———–(II)

On multiplying equations (i) and (ii), we obtain

AQ/4 . AQ/9 = cot θ. tanθ

AQ2/36 = 1

AQ2 = 36

AQ = √36 = ±6

However, height cannot be negative.

Therefore, the height of the tower is 6 m.


 

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