A) \[(\sqrt{3}-1)\,km\]
B) \[(\sqrt{3}+1)\,km\]
C) \[(\sqrt{3}-2)\,km\]
D) \[(\sqrt{3}+2)\,km\]
Correct Answer: B
Solution :
[b] Let \[AB=h\,km\] \[\therefore \]In \[\Delta {\mathrm O}\Alpha \Beta ,\]\[\tan 45{}^\circ =\frac{AB}{OB};\]\[OB=h\,km\] In \[\Delta {\mathrm O}L\Mu ,\]\[OM=2\cos 30{}^\circ =\sqrt{3}\,km\] \[\therefore \] \[LN=BM=(h-\sqrt{3})\,KM\] In \[\Delta \,OLM,\]\[\sin 30{}^\circ =\frac{AN}{LM}\] \[LM=2\sin 30{}^\circ =1\,km\] \[\therefore \] \[BN+LM+1km\] In \[\Delta \Alpha LN,\]\[\tan 60{}^\circ =\frac{AN}{LN}\] \[\sqrt{3}=\frac{AB=BN}{LN};\]\[\sqrt{3}=\frac{h-1}{h-\sqrt{3}}\] \[\sqrt{3}h-3=h-1\] \[=h\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3+1}}\] \[h=(\sqrt{3}+1)\,km\] |
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