The angle of elevation of a cloud from a point of \[200{\text{ }}metres\] above a lake is \[{30^o}\] and the angle of depression of the reflection of the cloud in the lake is \[{60^o}\] . Find the height of the cloud.
A.The height of cloud = 400m
B.The height of cloud = 280m
C.The height of cloud = 340m
D.None of these
Answer
181.5k+ views
Hint: We have given the angle of elevation cloud from the point \[200{\text{ }}metres\]above the lake is \[{30^o}\]an angle of depression is of reflection of the cloud is \[{60^o}\]. So, firstly we have to draw a rough diagram of the situation. After this, we calculate the base length. This base length helps us to find the height of the cloud from the ground level.
Complete step-by-step answer:
Thus is the diagram of the given situation;
The point is \[60m\] above the ground level. We take it as AB.
\[AB{\text{ }} = {\text{ }}CD\]
\[CD{\text{ }} = {\text{ }}200m\]
Let \[DE{\text{ }} = {\text{ }}h\]
So the height of the cloud \[ = {\text{ }}\left( {200 + h} \right){\text{ }}metre\] . If we calculate the value of h we can find the height of the cloud.
Let BD = \[x{\text{ }}metre\] .
Let the reflection of the cloud is at point F. As we know that the distance of the reflection from the ground level is the same as the distance of the object from the ground level. So,
\[CF{\text{ }} = {\text{ }}200 + h\]
Now from the triangle BDE \[\angle D{\text{ }} = {\text{ }}{90^o}\]
So triangle BDE is the right-angled triangle; $ \tan \theta = \dfrac{{perpendicular}}{{base}} $
∴ \[tan{30^o}\] $ = \dfrac{h}{x} $
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x} \Rightarrow x = \sqrt 3 h $ ……………..(i)
In \[\Delta {\text{ }}BDF\] , $ \angle D = {90^o} $
∴ triangle BDF is the right-angled triangle, therefore \[tan{60^o}\] $ = \dfrac{{DF}}{{BD}} $
$ \Rightarrow \sqrt 3 = \dfrac{{200 + 200 + h}}{x} $
$ \sqrt 3 x = 400 + h $ ………….(ii)
From the equations (i) value of \[x{\text{ }} = \] $ \sqrt 3 h $
Therefore, in (ii)
$ \sqrt 3 \left( {\sqrt 3 h} \right) = 400 + h $
$ 3h = 400 + h $
\[3h{\text{ }}-h{\text{ }} = {\text{ }}400\]
\[2h{\text{ }} = {\text{ }}400\]
$ h = \dfrac{{400}}{2} $
\[ = {\text{ }}200m\]
So, the value of h \[ = {\text{ }}200m\]
Now we can find the height of cloud
Height of cloud \[ = {\text{ }}200 + h{\text{ }} = {\text{ }}200 + 200{\text{ }} = {\text{ }}400m\]
So, option A is correct.
So, the correct answer is “Option A”.
Note: I.Angle of depression is a downward angle from the horizontal to the line of sight from the observer to some point of interest.
II.The tangent of the angle is equal to the side opposite to acute angle divided by the adjacent side of acute angle.
III.Angle of elevation is the upward angle from the horizontal line opf the sight from the observer to to some point of interest
Complete step-by-step answer:
Thus is the diagram of the given situation;

The point is \[60m\] above the ground level. We take it as AB.
\[AB{\text{ }} = {\text{ }}CD\]
\[CD{\text{ }} = {\text{ }}200m\]
Let \[DE{\text{ }} = {\text{ }}h\]
So the height of the cloud \[ = {\text{ }}\left( {200 + h} \right){\text{ }}metre\] . If we calculate the value of h we can find the height of the cloud.
Let BD = \[x{\text{ }}metre\] .
Let the reflection of the cloud is at point F. As we know that the distance of the reflection from the ground level is the same as the distance of the object from the ground level. So,
\[CF{\text{ }} = {\text{ }}200 + h\]
Now from the triangle BDE \[\angle D{\text{ }} = {\text{ }}{90^o}\]
So triangle BDE is the right-angled triangle; $ \tan \theta = \dfrac{{perpendicular}}{{base}} $
∴ \[tan{30^o}\] $ = \dfrac{h}{x} $
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x} \Rightarrow x = \sqrt 3 h $ ……………..(i)
In \[\Delta {\text{ }}BDF\] , $ \angle D = {90^o} $
∴ triangle BDF is the right-angled triangle, therefore \[tan{60^o}\] $ = \dfrac{{DF}}{{BD}} $
$ \Rightarrow \sqrt 3 = \dfrac{{200 + 200 + h}}{x} $
$ \sqrt 3 x = 400 + h $ ………….(ii)
From the equations (i) value of \[x{\text{ }} = \] $ \sqrt 3 h $
Therefore, in (ii)
$ \sqrt 3 \left( {\sqrt 3 h} \right) = 400 + h $
$ 3h = 400 + h $
\[3h{\text{ }}-h{\text{ }} = {\text{ }}400\]
\[2h{\text{ }} = {\text{ }}400\]
$ h = \dfrac{{400}}{2} $
\[ = {\text{ }}200m\]
So, the value of h \[ = {\text{ }}200m\]
Now we can find the height of cloud
Height of cloud \[ = {\text{ }}200 + h{\text{ }} = {\text{ }}200 + 200{\text{ }} = {\text{ }}400m\]
So, option A is correct.
So, the correct answer is “Option A”.
Note: I.Angle of depression is a downward angle from the horizontal to the line of sight from the observer to some point of interest.
II.The tangent of the angle is equal to the side opposite to acute angle divided by the adjacent side of acute angle.
III.Angle of elevation is the upward angle from the horizontal line opf the sight from the observer to to some point of interest
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
