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Question: State True or False: If the angle of elevation of a cloud from a point \(h\) meters above a lake ...

State True or False:
If the angle of elevation of a cloud from a point hh meters above a lake has measure α\alpha and the angle of depression of its reflection in the lake has measure β\beta , then height of the cloud is h(tanα+tanα)tanβtanβ\dfrac{{h(\tan \alpha + \tan \alpha )}}{{\tan \beta - \tan \beta }}.
A.True
B.False

Explanation

Solution

We will first draw the diagram using the given information and mark the positions of the cloud and its reflection with respect to the lake. Then we will apply the trigonometric ratio to find appropriate relations between the angles α\alpha and β\beta . Finally, we will find the height of the cloud in terms of the angles α\alpha and β\beta , then compare it with the given height to state whether the statement is true or false.

Formula used:
If α\alpha is an acute angle in a right-triangle, then tanα=opposite sideadjacent side\tan \alpha = \dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}

Complete step-by-step answer:
Let PP be the position of the cloud and PP' be the position of the reflection of the cloud.
Let OO and QQ be the points hh meters above the lake. Let the cloud be xx meters above the lake.
Let SRSR be the surface of the lake such that OS=QR=hOS = QR = h and PR=PR=xPR = P'R = x. We have to find xx.

From the figure,
PR=PQ+QRPR = PQ + QR
Substituting QR=hQR = h in the above equation, we get
\Rightarrow x = PQ + h \\\ \Rightarrow PQ = x - h \\\
Also, PQ=PR+RQ=x+hP'Q = P'R + RQ = x + h
Now in OPQ\vartriangle OPQ,
tanα=PQOQ\tan \alpha = \dfrac{{PQ}}{{OQ}}
Substituting PQ=xhPQ = x - h in the above equation, we get
tanα=xhOQ\Rightarrow \tan \alpha = \dfrac{{x - h}}{{OQ}}
Writing the above relation in terms of OQOQ, we get
OQ=xhtanαOQ = \dfrac{{x - h}}{{\tan \alpha }} ……………..(1)\left( 1 \right)
Now, consider OPQ\vartriangle OP'Q
tanβ=PQOQ\tan \beta = \dfrac{{P'Q}}{{OQ}}
tanβ=x+hOQ\Rightarrow \tan \beta = \dfrac{{x + h}}{{OQ}}
On cross multiplication, we get
OQ=x+htanβOQ = \dfrac{{x + h}}{{\tan \beta }} ……………..(2)\left( 2 \right)
Now equating equation (1)\left( 1 \right) and (2)\left( 2 \right), we get
xhtanα=x+htanβ\dfrac{{x - h}}{{\tan \alpha }} = \dfrac{{x + h}}{{\tan \beta }}
On Cross multiplication, we get
\Rightarrow (xh)tanβ=(x+h)tanα(x - h)\tan \beta = (x + h)\tan \alpha
Multiplying the terms on the LHS and RHS, we have
\Rightarrow xtanβhtanβ=xtanα+htanαx\tan \beta - h\tan \beta = x\tan \alpha + h\tan \alpha
Collecting the like terms on the LHS and RHS,
\Rightarrow xtanβxtanα=htanα+htanβx\tan \beta - x\tan \alpha = h\tan \alpha + h\tan \beta
Taking xx common from the LHS and hh common from the RHS, we get
\Rightarrow x(tanβtanα)=h(tanα+tanβ)x(\tan \beta - \tan \alpha ) = h(\tan \alpha + \tan \beta )
Taking (tanβtanα)(\tan \beta - \tan \alpha ) to the denominator in the RHS, we get
\Rightarrow x=h(tanα+tanβ)tanβtanαx = \dfrac{{h(\tan \alpha + \tan \beta )}}{{\tan \beta - \tan \alpha }}
Thus, the height of the cloud is h(tanα+tanβ)tanβtanα\dfrac{{h(\tan \alpha + \tan \beta )}}{{\tan \beta - \tan \alpha }}.
But we are given height of the cloud as h(tanα+tanα)tanβtanβ\dfrac{{h(\tan \alpha + \tan \alpha )}}{{\tan \beta - \tan \beta }}, which is not the same as what we have found.
Hence, the given statement is False.

Note: If a person stands and looks up at an object, the angle of elevation is the angle between the horizontal line of sight and the object. If a person stands and looks down at an object, the angle of depression is the angle between the horizontal line of sight and the object. Here, it is important to draw the diagram so that we can easily find the length of each side of the triangle.