The angle of elevation of a tower from a point 300 m above a lake is \[30{}^\circ .\] and the angle of depression of its reflection in the lake is \[60{}^\circ .\] Find the height of the tower. |
A) 600 m
B) 450 m
C) 200 m
D) 750 m
Correct Answer: A
Solution :
Let BC be the tower and E the point of observation 300 m above the lake surface. |
Draw \[AE\bot BC.\]BC' is the reflection of the tower BC in the lake such that |
\[BC=BC'=h\,\,m\] |
\[DE=AB=300\,\,m,\] |
\[\angle AEC=30{}^\circ \]and \[\angle AEC'=60{}^\circ \] |
\[AE=BD=x\,\,m\] |
\[AC=BC-BA\] |
\[=(h-300)\,\,m\] |
\[AC'=BC'+BA\] |
\[=(h+300)\,\,m\] |
In right \[\Delta CAE,\]\[\tan 30{}^\circ =\frac{AC}{AE}\] |
\[\frac{1}{\sqrt{3}}=\frac{h-300}{x}\] (i) |
In right \[\Delta CAE,\]\[\tan 60{}^\circ =\frac{AC'}{AE}\] |
\[\sqrt{3}=\frac{h+300}{x}\] (ii) |
On dividing, \[\frac{h-300}{h+300}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}\] |
\[\Rightarrow \]\[3\,\,(h-300)=(h+300)\]\[\Rightarrow \]\[h=600\,\,m\] |
Hence, the height of the tower is \[600\,\,m.\] |
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