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.

 

In above figure point O is an eye of observer and point P is an object so line of joining both point OP is a line of sight.

∠POA is the angle of elevation.

∠BPO is the angle of depression.

 

EXAMPLE

 The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30^\circ than when it is60^\circ . Find the height of the tower.

Solution:

Let AB be the tower and BC be the length of its shadow when sun’s altitude (angle of elevation from the top of the tower to the tip of the shadow) is 60^\circ and DB be the length of the shadow when the angle of elevation is30^\circ .

AB = h m, BC = x m

DB = (40 +x) m

In right triangle ABC,

Tan 60^\circ = AB/BC

\sqrt 3 = h/x

h = \sqrt 3 x………. (i)

In right triangle ABD,

Tan 30^\circ = AB/BD

1/\sqrt 3 =h/(x + 40) …….. (ii)

From (i) and (ii),

X (\sqrt 3 ) (\sqrt 3 ) = x + 40

3x = x + 40

2x = 40

x = 20

Substituting x = 20 in (i),

h = 20\sqrt 3

Hence, the height of the tower is 20\sqrt 3 m

 

EXAMPLE 2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30^\circ with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

Let AC be the broken part of the tree.

BC = 8 m

To Find: Height of the tree, which is AB

Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ABC,

Using Cosine and tangent angles,

Cos 30^\circ = BC/AC

We know that, Cos 30^\circ = \sqrt 3 /2

\sqrt 3 /2 = 8/AC

AC = 16/\sqrt 3 …(1)

Also,

Tan 30^\circ = AB/BC

1/\sqrt 3 = AB/8

AB = 8/\sqrt 3 ….(2)

From (1) and (2),

Total height of the tree = AB + AC = 16/\sqrt 3 + 8/\sqrt 3 = 24/\sqrt 3 = 8\sqrt 3 sm.

 

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