question_answer

The angles of elevation of the top of a vertical tower from two points. 30 metres apart, and on the same straight line passing through the base of tower, are \[30{}^\circ \] and \[60{}^\circ \] respectively. The height of the tower is

A) 10 m                    

B) 15 m

C) \[15\sqrt{3}\,\,m\]

D) 30 m

Correct Answer: C

Solution :

 Let AB is tower, the angles of elevation of the top of a vertical tower AB from two points D and C are \[{{30}^{o}}\] and \[{{60}^{o}}\] respectively. \[CD=30\]metre Let the height of tower be h metres and         \[BC=x\text{ }m\] In triangle \[ABC,\frac{AB}{BC}=\tan {{60}^{o}}\]                 \[\frac{h}{x}=\sqrt{3}\]                                 ?..(i) In triangle \[ABD,\frac{AB}{BD}=\tan {{30}^{o}}\] \[\therefore \]  \[\frac{h}{(x+30)}=\frac{1}{\sqrt{3}}\] or            \[\sqrt{3h}=x+30\] Put value of x from equation (i) in equation (ii), we get                 \[\sqrt{3h}=\frac{h}{\sqrt{3}}+30\] or            \[\sqrt{3h}-\frac{h}{\sqrt{3}}=30\] or            \[\frac{2h}{\sqrt{3}}=30\] \[\therefore \]  \[h=\frac{30\sqrt{3}}{2}=15\sqrt{3}\] metres So, height of the tower is \[15\sqrt{3}\] metre.