A) \[9(1+\sqrt{3})\]
B) \[\frac{9}{2}(1\sqrt{3}-1)\]
C) \[18(1+\sqrt{3})\]
D) \[18(\sqrt{3}-1)\]
Correct Answer: A
Solution :
Given, \[\angle HOA={{30}^{o}}\]and \[\angle HOB={{45}^{o}}.\]Let \[OH=h.\]then \[\tan (\angle HOA)=\frac{HA}{h}\]or \[HA=h\times \tan {{30}^{o}}\] \[\tan (\angle HOB)=\frac{HB}{h}\]or \[HB=h\times \tan {{45}^{o}}\] It takes 18 mins to travel distance AB, hence speed \[=\frac{HB-HA}{18}=\frac{h-(h/\sqrt{3})}{18}\] Time taken to travel \[HA=\frac{HA}{speed}=\frac{\frac{h}{\sqrt{3}}}{\frac{(h-\frac{h}{\sqrt{3}})}{18}}\] \[=\frac{18\times \sqrt{3}}{\sqrt{3}(\sqrt{3}-1)}\] \[=9(\sqrt{3}+1)\] Option A correct.You need to login to perform this action.
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