The angles of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively \[15{}^\circ \]and \[30{}^\circ .\] If A and B are on the same side of the tower and\[AB=48\,\,m,\]then the height of the tower is [SSC (FCI) 2012] |
A) \[24\sqrt{3}\,\,m\]
B) \[24\,\,m\]
C) \[24\sqrt{2}\,\,m\]
D) \[96\,\,m\]
Correct Answer: B
Solution :
Let height of the tower be \[h\,\,m\] and \[BC=x\,\,m\] |
In \[\Delta ACD,\] |
\[\tan 15{}^\circ =\frac{h}{x+48}\] |
\[\Rightarrow \]\[\tan \,\,(45{}^\circ -30{}^\circ )=\frac{h}{x+48}\] |
By the formula, |
\[\frac{\tan 45{}^\circ -\tan 30{}^\circ }{1+\tan 45{}^\circ \times \tan 30{}^\circ }=\frac{h}{x+48}\] |
\[\Rightarrow \] \[\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}=\frac{h}{x+48}\] |
\[\Rightarrow \] \[\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{h}{x+48}\] |
\[\Rightarrow \] \[\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{h}{x+48}\] |
\[\Rightarrow \] \[\frac{2\,\,(2-\sqrt{3})}{2}=\frac{h}{x+48}\] |
\[\therefore \] \[2-\sqrt{3}=\frac{h}{x+48}\] (i) |
In \[\Delta BCD,\] \[\tan 30{}^\circ =\frac{h}{x}\] |
\[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{h}{x}\] |
\[\Rightarrow \] \[\sqrt{3}n=x\] (ii) |
From Eq. (i), |
\[2-\sqrt{3}=\frac{h}{\sqrt{3}h+48}\] |
\[\Rightarrow \]\[2\sqrt{3}h-3h+(2-\sqrt{3})48=h\] |
\[\Rightarrow \] \[h+3h-2\sqrt{3}h=(2-\sqrt{3})\times 48\] |
\[\Rightarrow \] \[4h-2\sqrt{3}h=(2-\sqrt{3})\times 48\] |
\[\Rightarrow \] \[2h\,\,(2-\sqrt{3})=(2-\sqrt{3})\times 48\] |
\[\Rightarrow \] \[2h=48\] |
\[\therefore \] \[h=24\,\,m\] |
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