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Question: The angle of elevation of a chord from a point ‘h’ m above the lake is\[\alpha \]and angle of depres...

The angle of elevation of a chord from a point ‘h’ m above the lake isα\alpha and angle of depression of reflection in the lake isβ\beta . Prove that the height of the cloud from surface of water is
h(tan⁡β+tan⁡α)tan⁡β−tan⁡α\dfrac{h\left( \tan \beta +\tan \alpha \right)}{\tan \beta -\tan \alpha }.

Explanation

Solution

Hint: Draw the figure as per mentioned in the question. The angle of elevation of cloud is at a height h and 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. Consider the triangles formed and solve it to get an expression for x.

Complete step-by-step answer:

Let us assume that p is the point which is at ‘h’ meters distance from the lake is taken as C’. Refer to the figure.

Let us assume that the distance of the cloud from the lake is ‘x’ meters.
\therefore $$$$\alpha represents the angle of elevation of cloud above the lake andβ\beta represents the angle of depression of reflection of cloud in the lake.
We can say that,
BC=BC′=xBC=B{{C}^{'}}=x.
From the figure we can make out that AP = BM = h.
∴\therefore The length is
C′M=BC′+MB=x+h{{C}^{'}}M=B{{C}^{'}}+MB=x+h.
Let us consider,△CPM\vartriangle CPM where∠CPM=α\angle CPM=\alpha and right angled at M, so△CPM\vartriangle CPM is a right angle triangle. By basic trigonometry,
tan⁡α\tan \alpha = opposite side/ adjacent side =CMPM\dfrac{CM}{PM}.
⇒tan⁡α=CMPM⇒PM=CMtan⁡α=x−htan⁡α\Rightarrow \tan \alpha =\dfrac{CM}{PM}\Rightarrow PM=\dfrac{CM}{\tan \alpha }=\dfrac{x-h}{\tan \alpha }
CB=CM+MBCB=CM+MB
CM=x−hCM=x-h
∴PM=x−htan⁡α−(1)\therefore PM=\dfrac{x-h}{\tan \alpha }-(1)
Now consider,△PMC′\vartriangle PM{{C}^{'}}
tan⁡β=CM′PM\tan \beta =\dfrac{C{{M}^{'}}}{PM}
⇒PM=x+htan⁡β−(2)\Rightarrow PM=\dfrac{x+h}{\tan \beta }-(2)[From figure]
Equate both equation (1) and (2) and cross multiplying.
x+htan⁡α=x+htan⁡β\dfrac{x+h}{\tan \alpha }=\dfrac{x+h}{\tan \beta }
tan⁡β(x−h)=(x+h)tan⁡α\tan \beta \left( x-h \right)=\left( x+h \right)\tan \alpha
⇒xtan⁡β−htan⁡β=xtan⁡α+htan⁡α\Rightarrow x\tan \beta -h\tan \beta =x\tan \alpha +h\tan \alpha
h(tan⁡α+tan⁡β)=x(tan⁡β−tan⁡α)h\left( \tan \alpha +\tan \beta \right)=x\left( \tan \beta -\tan \alpha \right)
∴x=h(tan⁡α+tan⁡β)tan⁡β−tan⁡α\therefore x=\dfrac{h\left( \tan \alpha +\tan \beta \right)}{\tan \beta -\tan \alpha }
Hence proved.

Note: Remember to take the value of the height of the cloud from the lake i.e. elevation of cloud and depression of reflection of cloud as x. So it makesBC=BC′=xBC=B{{C}^{'}}=x. Solution of the two triangles formed from the surface of the lake to the elevation and reflection will give us the value of ‘x’.