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