Question

The angle of elevation of a cloud from point h meter above the lake is \[\alpha \] and the angle of depression of its reflection in the lake is \[\beta \], prove that the distance of the cloud from the point of observation is

\[\dfrac{2h\sec \alpha }{\tan \beta -\tan \alpha }\].

Answer 1

Hint: Draw the figure as per mentioned in the question. The angle is elevation of cloud is at a height h above the lake. Take the height of the cloud from the lake as x. The height of elevation and depression of reflection of clouds is the same.

Complete step-by-step answer:

Let us assume that P is the point which is at h meter distance from the lake. C is taken as the position of cloud. Let C’ be the reflection of the cloud in the lake. Refer to the figure.

Let us also assume that the distance of the cloud from the lake is ‘x’ meters.

Let \[\alpha \] represent the angle of elevation of cloud above the lake and \[\beta \] represents the angle of depression of reflection of cloud in the lake.

We can say that, \[BC=BC'=x\].

From the figure we can say that, AP = BQ = h.

\[\therefore \] The length of \[C'Q=BC'+QB=x+h\].

Now let us consider, \[\Delta PQC\].

Here, \[\angle CPQ=\alpha \] and right angled are Q, so \[\Delta PQC\] is a right angled triangle by basic trigonometry.

\[\sin \alpha \] = opposite side/ hypotenuse\[=\dfrac{CQ}{PC}\].

\[  \therefore \sin \alpha =\dfrac{x}{PC} \]

\[ \Rightarrow PC=\dfrac{x}{\sin \alpha } \]

\[ \therefore x=CP\sin \alpha -(1) \]

Now, \[\tan \alpha \] = opposite side / adjacent side \[=\dfrac{CQ}{PQ}\].

\[ \therefore \tan \alpha =\dfrac{x}{PQ} \]

\[ \therefore x=PQ\tan \alpha -(2) \]

Now let us consider the right triangle PQC’.

\[\tan \beta \] = opposite side/ adjacent side \[=\dfrac{QC'}{PQ}=\dfrac{BQ+BC'}{PQ}\]

\[\therefore \tan \beta =\dfrac{h+x+h}{PQ}\] 

\[ \because BC'=x+h \]

\[ \because BQ=h \]

\[  \tan \beta =\dfrac{2h+x}{PQ} \]

\[  \therefore PQ=\dfrac{2h+x}{\tan \beta }-(3) \]

From (2), \[PQ=\dfrac{x}{\tan x}\]

Now let us compare equation (2) and equation (3).

\[\dfrac{x}{\tan x}=\dfrac{2h+x}{\tan \beta }\]

Cross multiply and simplify the above expression.

\[ x\tan \beta =\left( 2h+x \right)\tan \alpha \]

\[ x\tan \beta =2h\tan \alpha +x\tan \alpha \]

\[ x\tan \beta -x\tan \alpha =2h\tan \alpha \]

\[ x\left( \tan \beta -\tan \alpha \right)=2h\tan \alpha \]

\[ \therefore x=\dfrac{2h\tan \alpha }{\tan \beta -\tan \alpha }-(4) \]

Now let us compare equation (1) and equation (4).

\[ CP\sin \alpha =\dfrac{2h\tan \alpha }{\tan \beta -\tan \alpha } \]

\[ \therefore CP=\dfrac{2h\tan \alpha }{\sin \alpha \left( \tan \beta -\tan \alpha \right)} \]

We know, \[\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }\].

\[\therefore CP=\dfrac{2h}{\sin \alpha }\times \dfrac{\sin \alpha }{\cos \alpha }\times \dfrac{1}{\left( \tan \beta -\tan \alpha \right)}\] (\[\because \] Cancel out \[\sin \alpha \] from numerator and denominator)

\[ CP=\dfrac{2h}{\cos \alpha }\times \dfrac{1}{\left( \tan \beta -\tan \alpha \right)} \]

\[ CP=\dfrac{2h\sec \alpha }{\tan \beta -\tan \alpha } \]

Hence we found the distance of the cloud from the point of observer is \[\dfrac{2h\sec \alpha }{\tan \beta -\tan \alpha }\].

Note: Remember to take the value of height of the cloud from the lake i.e. elevation of cloud and depression of reflection of cloud as x. So it makes BC = BC’ = x. Solution of the triangles formed from the surface of the lake to the elevation and reflection will give value of x.