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.You need to login to perform this action.
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