question_answer

. Angles of elevation of the top of a tower from three point (collinear) A, Band C on a road leading to the foot of the tower are \[{{30}^{o}},{{45}^{o}}\]and \[{{60}^{o}}\]respectively. The ration of AB to BC is

A) \[\sqrt{3}:1\]                    

B) \[\sqrt{3}:2\]

C)  \[1:2\]                                 

D) \[2:\sqrt{3}\]

Correct Answer: A

Solution :

By sin law, in \[\Delta \,ABQ\] \[\frac{\sin 15{}^\circ }{AB}=\frac{\sin 30{}^\circ }{BQ}\]  \[\Rightarrow \]              \[BQ=\frac{\sin 30{}^\circ .\,AB}{\sin 15{}^\circ }\]           ?(i) By sine law, in \[\Delta BCQ\] \[\frac{\sin 120{}^\circ }{BQ}=\frac{\sin 15{}^\circ }{BC}\] \[\Rightarrow \]               \[BQ=\frac{BC\sin 120{}^\circ }{\sin 15{}^\circ }\]              ?(ii) From Eqs. (i) and (ii), we get \[\frac{\sin 30{}^\circ \cdot AB}{\sin 15{}^\circ }=\frac{BC\cdot \sin 120{}^\circ }{\sin 15{}^\circ }\] \[\Rightarrow \]               \[\frac{AB}{BC}=\frac{\sin 120{}^\circ }{\sin 30{}^\circ }=\frac{\cos 30{}^\circ }{\sin 30{}^\circ }=\frac{\sqrt{3}/2}{1/2}\] \[AB:BC=\sqrt{3}:1\]