Answer
Verified
410.1k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Advantages and disadvantages of science
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
10 examples of evaporation in daily life with explanations
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
How do you graph the function fx 4x class 9 maths CBSE
Difference Between Plant Cell and Animal Cell