A) \[(3+\sqrt{3})\,km\]
B) \[(3-\sqrt{3})\,km\]
C) \[2\sqrt{3}\,km\]
D) \[3\sqrt{3}\,km\]
Correct Answer: B
Solution :
Let \[PQ=h\,\,m\] be a plane From right angled \[\Delta \,PAQ,\] \[\tan {{45}^{o}}=\frac{PQ}{AQ}\] \[\therefore \] \[1=\frac{h}{x}\] or \[x=h\] ??(i) Again from right angled \[\Delta PBQ,\] © \[\tan {{60}^{o}}=\frac{PQ}{QB}\] or \[\sqrt{3}=\frac{h}{2-x}\] From equation (i), \[\sqrt{3}=\frac{h}{2-h}\] or \[2\sqrt{3}-\sqrt{3}\,h=h\] or \[h+\sqrt{3}h=2\sqrt{3}\] or \[h(1+\sqrt{3})=2\sqrt{3}\] or \[h=\frac{2\sqrt{3}}{1+\sqrt{3}}=\frac{2\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}\] \[=\frac{2\sqrt{3}-6}{1-3}=\frac{-2(3-\sqrt{3})}{-2}\] \[=(3-\sqrt{3})km\]You need to login to perform this action.
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