Some Applications of Trigonometry
Trigonometry can be used in many ways in the things around us like we can use it for calculating the height and distance of some objects without calculating them actually.
Heights and Distances
To find the height of an object or to find the distance of an object to the other we must know the meaning of some points -
Line of Sight
When we look at some object then the line made by our vision to the object is called the Line of Sight.
Horizontal Line
A horizontal line is the distance between the observer and the object.
Angle of Elevation
An angle of elevation is the angle made by the line of sight to the top of the object and the horizontal line. It is above the horizontal line i.e. the angle of elevation is made when we look up to the object.
Angle of Depression
An angle of depression is made when the observer needs to look down to see the object.The angle between the horizontal line and the line of sight is the angle of depression when the horizontal line is above the angle.
Some Solved Examples
To solve the problems related to the angle of elevation and angle of depression we must remember trigonometric ratios, trigonometric table and the trigonometric identities.
Example: 1
Find the height of the flagpole if the angle of elevation is 30° and the distance of flag from the observer is 15 m.
Solution:
Let x be the height of the flagpole.
Horizontal line is 15m.
Now, to calculate x, we need to take the trigonometric ratio which includes perpendicular and base.
Hence, the flagpole is 9 m.
Example: 2
A child was playing at the top of the hill. He had thrown a stone in the lake from the hill, the distance covered by the stone was 150 m and the angle of depression was 30°, then what is the height of the hill?
Solution:
Let the height of the hill be h.
The hypotenuse is 150 m and the angle of depression is 30°.
Hence the height of the hill is 75 m.
Example: 3
A girl who is 1.2 m tall is watching a balloon moving in a horizontal line at a height of 88.2 m from the ground.The angle of elevation from her eyes is changed from 60° to 30°. Calculate the distance travelled by balloon.
Solution:
In ΔACE
In ΔBCG
CG = 87√3 m
Distance travelled by balloon = EG = CG - CE
87√3 - 29√3 = 58√3 m