Answer:
53.33 meters
Step-by-step explanation:
Let AB represents the height of the cliff,
( where, A is top and B is bottom ),
Also, C and D represents the shadow of the cloud and cloud in the sky respectively,
Suppose E is a point in the segment CD,
Such that,
AB = DE = 40 meters,
According to the question,
![m\angle CAE = 30^{\circ}](https://tex.z-dn.net/?f=m%5Cangle%20CAE%20%3D%2030%5E%7B%5Ccirc%7D)
![m\angle EAD = 60^{\circ}](https://tex.z-dn.net/?f=m%5Cangle%20EAD%20%3D%2060%5E%7B%5Ccirc%7D)
Since,
![\tan =\frac{\text{Perpendicular}}{\text{Base}}](https://tex.z-dn.net/?f=%5Ctan%20%3D%5Cfrac%7B%5Ctext%7BPerpendicular%7D%7D%7B%5Ctext%7BBase%7D%7D)
![\implies \tan 60^{\circ}=\frac{DE}{AE}](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctan%2060%5E%7B%5Ccirc%7D%3D%5Cfrac%7BDE%7D%7BAE%7D)
![\sqrt{3}=\frac{40}{AE}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%3D%5Cfrac%7B40%7D%7BAE%7D)
![\implies AE = \frac{40}{\sqrt{3}}](https://tex.z-dn.net/?f=%5Cimplies%20AE%20%3D%20%5Cfrac%7B40%7D%7B%5Csqrt%7B3%7D%7D)
Now,
![\tan 30^{\circ}=\frac{CE}{AE}](https://tex.z-dn.net/?f=%5Ctan%2030%5E%7B%5Ccirc%7D%3D%5Cfrac%7BCE%7D%7BAE%7D)
![\frac{1}{\sqrt{3}}=\frac{\sqrt{3}CE}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%3D%5Cfrac%7B%5Csqrt%7B3%7DCE%7D%7B40%7D)
![\implies CE = \frac{40}{3}](https://tex.z-dn.net/?f=%5Cimplies%20CE%20%3D%20%5Cfrac%7B40%7D%7B3%7D)
Hence,
The height of the cloud above the lake = CE + ED
![=\frac{40}{3}+40=13.33+40 = 53.33\text{ meters}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40%7D%7B3%7D%2B40%3D13.33%2B40%20%3D%2053.33%5Ctext%7B%20meters%7D)