question_answer

The angular elevation of a tower CD at a point A due south of it is\[{{60}^{o}}\]and at a point B due west of A, the elevation is\[{{30}^{o}}\]. If AB = 3 km, the height of the tower is [MP PET 1998]

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