A) \[15\sqrt{3}m\]
B) \[\frac{15}{\sqrt{3}}m\]
C) \[15m\]
D) \[20m\]
Correct Answer: A
Solution :
\[\tan 60{}^\circ =\frac{h}{x}\Rightarrow \frac{\sqrt{3}}{1}=\frac{h}{x}\Rightarrow h=\sqrt{3}x\] .....(i) \[\tan 30{}^\circ =\frac{h}{60-x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{60-x}\Rightarrow 60-x=\sqrt{3}h\] .....(ii) From equation (i) and (ii), \[60-x=\sqrt{3}(\sqrt{3}x)\] \[\frac{60}{4}=x\Rightarrow x=15\] Then \[h=\sqrt{3}x\Rightarrow h=15\sqrt{3}\] metre.You need to login to perform this action.
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