A) \[\frac{5\sqrt{3}}{2}\,m\]
B) \[\frac{2\sqrt{3}}{5}\,m\]
C) \[\frac{2}{5\sqrt{3}}\,m\]
D) \[\frac{4}{5\sqrt{3}}\,m\]
Correct Answer: A
Solution :
[a] Let the height of the post \[=h\,\,m\] In \[\Delta ACB,\] \[\tan 30{}^\circ =\frac{h}{5+x}\] (i) In\[\Delta DAB,\]\[\tan 60{}^\circ =\frac{h}{x}\]\[\Rightarrow \]\[x=\frac{h}{\sqrt{3}}\] From Eq. (i), we get \[\frac{1}{\sqrt{3}}=\frac{h}{5+\frac{h}{\sqrt{3}}}\] \[\Rightarrow \] \[\frac{5}{\sqrt{3}}+\frac{h}{3}=h\]\[\Rightarrow \]\[h-\frac{h}{3}=\frac{5}{\sqrt{3}}\] \[\therefore \] \[h=\frac{5\times 3}{2\times \sqrt{3}}=\frac{5\sqrt{3}}{2}\,m\] |
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