## HEIGHTS AND DISTANCES

In this page heights and distances we are going to see the application problems on trigonometry.

Before going to example problems first let us see the two types of angles

• Angle of elevation
• Angle of depression

Angle of elevation:

If the object is above the horizontal level from the eye we have to lift up our head to view the object.In this process our eyes move though the angle.This angle is called angle of elevation. Angle of depression:

If the object is below the horizontal level from the eye,we have to move downwards our head to see the object.In this process our eyes move though an angle.This angle i called the angle of depression. Example 1:

A tree was observed from two points in a line with it but on opposite sides of it.The distance between the point is 120m. If the angle of elevation are 30° and 45° find the height of the tree. Here AB is the height of the tree..Let C and D be the two points in a line with it lying on opposite sides of the light house.At C ,the angle of elevation of A is 45°. That is ∠BCA = 45°. At D,the angle of elevation of A is 30°. That is ∠BDA = 30°

Now let us consider the triangle ACB.Here we need to find the height of AB.That is opposite side.But here we know the measurement of adjacent side.For that we have to use tan θ.Here θ is 45°.

tan θ = (Opposite side)/(Adjacent side)

tan 45° = AB/CB

1 = AB/x

x = AB

AB = x ----------(1)

Now let us consider the triangle ABD.Here we need to find the height of AB.That is opposite side.But here we know the measurement of adjacent side.For that we have to use tan θ.Here θ is 30°.

tan θ = (Opposite side)/(Adjacent side)

tan 30° = AB/BD

1/√3 = AB/(120-x)

(120-x) = AB√3

(120-x)/√3 = AB

AB = (120-x)/√3 -------(2)

(1) = (2)

x =  (120-x)/√3

√3x = 120- x

√3x + x = 120

x (√3 +1) = 120

x = 120/(√3 + 1)

x = 120/(1.732+1)

x = 120/2.732

x = 43.92 m

Height of the tree = 43.92 m

Related Topics

Heights And Distances to trigonometry 