question_answer

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),

\[2-\sqrt{3}=\frac{h}{\sqrt{3}h+48}\]

\[\Rightarrow \]\[2\sqrt{3}h-3h+(2-\sqrt{3})48=h\]

\[\Rightarrow \]               \[h+3h-2\sqrt{3}h=(2-\sqrt{3})\times 48\]

\[\Rightarrow \]               \[4h-2\sqrt{3}h=(2-\sqrt{3})\times 48\]

\[\Rightarrow \]               \[2h\,\,(2-\sqrt{3})=(2-\sqrt{3})\times 48\]

\[\Rightarrow \]                           \[2h=48\]

\[\therefore \]                              \[h=24\,\,m\]