question_answer

If the angles of elevation of a tower from two distant points a and \[b\,\,(a>b)\] from its foot and in the same straight line and on the same side of it, are \[30{}^\circ \] and \[60{}^\circ ,\] then the height of the tower is  [SSC (CPO) 2013]

A) \[\sqrt{\frac{a}{b}}\]                             

B) \[\sqrt{a+b}\]

C) \[\sqrt{ab}\]      

D) \[\sqrt{a-b}\]

Correct Answer: C

Solution :

In the above figure, there are two elevation angles \[60{}^\circ \]and \[30{}^\circ \]of a tower.

Let the height of the tower he h and distance points are a and \[b\,\,(a>b).\]

Now, in \[\Delta ACB\]

\[\tan 60{}^\circ =\frac{AB}{BC}=\frac{h}{b}\]\[\Rightarrow \]\[\sqrt{3}=\frac{h}{b}\]

\[h=b\sqrt{3}\]               (i)

Now, again in \[\Delta ADB\]

\[\tan 30{}^\circ =\frac{AB}{BD}=\frac{h}{a}\]\[\Rightarrow \]\[\frac{1}{\sqrt{3}}=\frac{h}{a}\]

\[\Rightarrow \]               \[h=\frac{a}{\sqrt{3}}\]                         (ii)

On multiplying Eqs. (i) and (ii), we get

\[{{h}^{2}}=(b\sqrt{3})\times \left( \frac{a}{\sqrt{3}} \right)\]\[\Rightarrow \]\[{{h}^{2}}=ab\]

\[\therefore \]      \[h=\sqrt{ab}\]