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, \[{{x}^{2}}+{{3}^{2}}={{y}^{2}}\] \[\Rightarrow \,\,{{\left( \frac{h}{\sqrt{3}} \right)}^{2}}+{{3}^{2}}={{(\sqrt{3}h)}^{2}}\Rightarrow h=\frac{3\sqrt{6}}{4}km\]You need to login to perform this action.
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