question_answer

The angle of elevation of a tower from m a point is  \[30{}^\circ \]. At a point on the horizontal line passing through the foot of the tower and 50 metres nearer it, the angle of elevation is \[60{}^\circ \]. The distance of the first point from the tower is

A)  50 metres       

B)         75 metres  

C)  100 metres       

D)         150 metres

Correct Answer: B

Solution :

 Let \[AC=x\text{ }m\] and \[AB=h\text{ }m\] From  \[\Delta s\,\,BCA,\]                                 \[\tan {{30}^{o}}=\frac{AB}{AC}\] or            \[\frac{1}{\sqrt{3}}=\frac{h}{x}\] or            \[h=\frac{x}{\sqrt{3}}\] From \[\Delta \,BDA\,\,\tan {{60}^{o}}=\frac{AB}{AD}\] or            \[\sqrt{3}=\frac{h}{x-50}\] or            \[h=(x-50)\,\sqrt{3}\] Equating the two values of h, we get                 \[\frac{x}{\sqrt{3}}=(x-50)\sqrt{3}\] or            \[x=(x-50)\,3=3x-150\] or            \[2x=150\] \[\therefore \] \[x=75\,m\]