Question

The angle of elevation of a cloud from a point 120m above a lake is 30° and the angle of

depression of its reflection in the lake is 60°.Find the height of the cloud.

Answer 1

Answer:

The answer is 240 m.

Step-by-step explanation:

Hope you get itThe height of cloud is 240 m

Step-by-step explanation:

Refer the attached figure

Let the height of cloud be H i.e. CE

Height = Length of image

So, CD =EF = H

AB = DE = 120 m

CD = CE - DE = H - 120

FD = FE+DE =H+120

In ΔBCD

 --- 1

In ΔBDF

--- 2

Equate 1 and 2

Hence The height of cloud is 240 m

Answer 2

Answer:

240m is the height of the cloud .

Step-by-step explanation:

Explanation:

Given , elevation of cloud from a point 120m above a lake is 30 °

angle of depression of its reflection in the lake is 60°

Here ,height = length of the image .

So , AB = DE = 120m .    

CD = CE - DE = H-120  ........(I)

FD = FE +DE = H+120.....(II)

Step 1:

In Δ BCD

$tan\theta = \frac{p }{b}$ (where p is perpendicular and b is base )

tan30 = $\frac{CD}{BD}$

⇒$\frac{1}{\sqrt{3} } = \frac{H-120}{BD}$ ⇒$BD = \frac{H-120}{\frac{1}{\sqrt{3} } }$ .......(II)

(CD = H-120 from equation i)

Similarly , in ΔBDF

$tan\theta = \frac{p }{b}$

⇒tan60 = $\frac{DF}{BD}$ = $\sqrt{3} = \frac{H+120}{BD}$

⇒BD = $\frac{H+120}{\sqrt{3} }$ .......(IV)

Step2:

Now from equation (iii) and (iv) we get

$\frac{H-120}{\frac{1}{\sqrt{3} } }$ = $\frac{H+120}{\sqrt{3} }$

⇒$\sqrt{3}(H-120 ) = (H+120)\frac{1}{\sqrt{3} }$

⇒ 3H - 360= H +120

⇒ 2H = 480

⇒H = 240m

Final answer :

Hence , the height of the cloud is 240m .