Question

1.From a point, 36 m above the surface of a
jake, the angle of elevation of a bird is
observed to be 30° and angle of depression of
its image in the water of the lake is observed
to be 60°. Find the actual height of the bird
above the surface of the lake.​

Answer 1

The actual height of the bird

above the surface of the lake is 48m.

• Let P be the point 36 m above the surface.

• L be the point on the lake(image of bird) at which angle of declination is 60°

• B be the point at which bird is present and angle of inclination is 30°

• Let M be a point on the line joining B and L such that PM is parallel to ground.

• So ML is also 36 m.

•Now, tan 60°= ML/PM

•√3 = 36/PM

• PM = 36/√3

• PM = 12√3

•Now, tan 30° = BM/PM

• 1/√3 = BM /(12√3)

• BM =12

•So total height = BM + ML

= 12 + 36

=48 m

Answer 2

Given :

From a point, 36 m above the surface of a  lake, the angle of elevation of a bird is  observed to be 30° and angle of depression of  its image in the water of the lake is observed  to be 60°.

To find :

Find the actual height of the bird  above the surface of the lake.​

Solution:

Refer the attached figure

Let A be the point 36 m above the surface of a  Lake

So, AC = DE = 36 m

let B be the position of bird and B' be the image of bird

The angle of elevation of a bird is  observed to be 30° i.e. ∠BAE =30°

The angle of depression of  its image in the water of the lake is observed  to be 60°i.e.∠B'AE =60°

Let BE be the h

Height of bird above surface = BD = BE+ED=h+36

Height of Bird above the surface = Length of image below the surface

So, BD = B'D

Now B'E = BD+DE=36+36+h=72+h

Now in triangle ABE

$\frac{Perpendicular}{Base} = Tan \theta \\\frac{BE}{AE} = tan 30^{\circ}\\\frac{h}{AE}=\frac{1}{\sqrt{3}}\\\sqrt{3}h=AE ---1$

In triangle AB'E

$\frac{B'E}{AE} = tan 30^{\circ}\\\frac{72+h}{AE}=\sqrt{3}\\\frac{72+h}{\sqrt{3}}=AE ---2$

Equate 1 and 2

$\sqrt{3}h=\frac{72+h}{\sqrt{3}}$

3h=72+h

2h=72

h=36

So, Height of bird above surface = BD = BE+ED=h+36= 36+36 = 72 m

Hence The Height of bird above surface is 72 m