A) \[a\sqrt{3}\]
B) \[\frac{a\sqrt{3}}{\sqrt{3}-1}\]
C) \[\frac{a(3+\sqrt{3})}{2}\]
D) \[a(\sqrt{3}-1)\]
Correct Answer: C
Solution :
In \[\Delta ABC,\] \[\tan 30{}^\circ =\frac{AC}{BC}\] or \[\frac{1}{\sqrt{3}}=\frac{x}{BC}\] |
or \[BC=x\sqrt{3}\]and in \[\Delta ADE,\] \[\tan 45{}^\circ =\frac{a+x}{DE}\] |
or \[1=\frac{a+x}{x\sqrt{3}}\] or \[x\sqrt{3}=a+x,\]\[\Rightarrow \,x(\sqrt{3}-1)=a\] |
or \[x=\frac{a}{\sqrt{3}-1}\] |
Therefore, height of the tower, |
\[a+x=a+\frac{a}{\sqrt{3}-1}\] |
\[=a\left[ \frac{\sqrt{3}-1+1}{\sqrt{3}-1} \right]=\frac{a\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{a(3+\sqrt{3})}{2}\] |
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