If the angle of elevation of a tower from a distance 100 m from its foot is \[30{}^\circ ,\] then the height of the tower is |
[SSC (CGL) Mains 2014] |
A) \[\frac{100}{\sqrt{3}}\,m\]
B) \[100\sqrt{3}\,m\]
C) \[\frac{50}{\sqrt{3}}\]
D) \[50\sqrt{3}\,m\]
Correct Answer: A
Solution :
Let the height of tower be h m. |
\[\tan 30{}^\circ =\frac{AB}{BC}\]\[\Rightarrow \]\[\frac{1}{\sqrt{3}}=\frac{h}{100}\] |
\[\Rightarrow \] \[h=\frac{100}{\sqrt{3}}m\] |
\[\therefore \] Height of the tower is \[\frac{100}{\sqrt{3}}m.\] |
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