question_answer

Find the angle of elevation of the top of a tower of height \[100\sqrt{3}m\] from a point at a distance of 100m from the foot of the tower on a horizontal plane.

A) \[{{45}^{o}}\]                        

B) \[{{60}^{o}}\]  

C) \[{{75}^{o}}\]            

D)        \[{{30}^{o}}\]

Correct Answer: B

Solution :

 Let\[\phi \] be the angle of elevation and \[AC=\] height of tower             \[=100\sqrt{3}m\] \[\therefore \]      \[AB=100\,\,m\]             \[\tan \phi =\frac{AC}{AB}=\frac{100\sqrt{3}}{100}=\sqrt{3}\] or         \[\tan \phi =\tan {{60}^{o}}\] \[\Rightarrow \]   \[\phi ={{60}^{o}}\] Hence angle of elevation is\[{{60}^{o}}\]