question_answer

The angle of elevation of a tower from a point on the ground is \[30{}^\circ \]. At a point on the horizontal line passing through the foot of the tower and 100 metres nearer to it. If the angle of elevation is found to be \[60{}^\circ \], then height of the tower is

A) \[50\sqrt{3}\text{ }meters\]

B) \[\frac{50}{\sqrt{3}}\text{ }meters\]

C) \[100\sqrt{3}\text{ }meters\]

D)   \[\frac{100}{\sqrt{3}}\text{ }meters\]  

Correct Answer: A

Solution :

 Let \[AB=h\] be the height of the tower, and let      \[AD=x\text{ }m\]                 From right angled \[\Delta \,ABD,\]                 \[\tan {{60}^{o}}=\frac{AB}{AD}\] or            \[\sqrt{3}=\frac{h}{x}\] or            \[x=\frac{h}{\sqrt{3}}\]                                 ?..(i) From right angled \[\Delta \,BCA,\]                 \[\tan {{30}^{o}}=\frac{AB}{CA}\] or            \[\frac{1}{\sqrt{3}}=\frac{h}{100+x}\] or            \[h=\frac{100+x}{\sqrt{3}}=\frac{100+\frac{h}{\sqrt{3}}}{\sqrt{3}}\] or            \[\sqrt{3}\,h=100+\frac{h}{\sqrt{3}}\] or            \[\frac{2}{\sqrt{3}}h=100\] \[\therefore \]  \[h=50\sqrt{3}\,m\]