question_answer

          The angular elevation of a tower CD at a point A due south of it is \[60{}^\circ \]and at a point B due west of A, the elevation is \[30{}^\circ \]. If \[AB=3\text{ }km,\]the height of the tower is

A) \[2\sqrt{3}\,km\]

B)               \[2\sqrt{6}\,km\]

C) \[\frac{3\sqrt{3}}{2}\,km\]          

D)   \[\frac{3\sqrt{6}}{4}\,km\]

Correct Answer: D

Solution :

From \[\Delta \,CDA,\] \[x=h\,\cot 60{}^\circ =\frac{h}{\sqrt{3}}\] From \[\Delta \,ABC,\] \[y=h\,\cot 30=h\sqrt{3}\] From \[\Delta ABC,\]by Pythagoras theorem, \[{{x}^{2}}+{{3}^{2}}={{y}^{2}}\] \[\Rightarrow \,\,{{\left( \frac{h}{\sqrt{3}} \right)}^{2}}+{{3}^{2}}={{(\sqrt{3}h)}^{2}}\Rightarrow h=\frac{3\sqrt{6}}{4}km\]