The length of the shadow of a tower standing on level ground is found to 2x meter longer when the sun’s elevation is 30° than when it was 45 °. The height of the tower in meters is.
A. $(\sqrt{3}+1) x$ B. $(\sqrt{3}-1) x$ C. $2 \sqrt{3} x$ D. $3 \sqrt{2} x$ Answer: Option AShow Answer
Solution(By Apex Team)
AB is a tower
BD and BC are its shadows and CD = 2x
$\begin{array}{l}\tan45^{\circ}=\frac{AB}{DB}\\
\Rightarrow1=\frac{h}{y}\Rightarrow y=h\\
\text{ and }\tan30^{\circ}=\frac{AB}{CB}\\
\begin{aligned}\Rightarrow\frac{1}{\sqrt{3}}=\frac{h}{2x+y}\\
\Rightarrow2x+y=\sqrt{3}h\\
\Rightarrow\sqrt{3}h-h=2x\\
\Rightarrow h(\sqrt{3}-1)=2x\\
\Rightarrow h=\frac{2x}{\sqrt{3}-1}\\
=\frac{2x(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\
=\frac{2x(\sqrt{3}+1)}{3-1}\\
=\frac{2x(\sqrt{3}+1)}{2}\\
=x(\sqrt{3}+1)\end{aligned}\end{array}$
$\begin{array}{l}\tan45^{\circ}=\frac{AB}{DB}\\
\Rightarrow1=\frac{h}{y}\Rightarrow y=h\\
\text{ and }\tan30^{\circ}=\frac{AB}{CB}\\
\begin{aligned}\Rightarrow\frac{1}{\sqrt{3}}=\frac{h}{2x+y}\\
\Rightarrow2x+y=\sqrt{3}h\\
\Rightarrow\sqrt{3}h-h=2x\\
\Rightarrow h(\sqrt{3}-1)=2x\\
\Rightarrow h=\frac{2x}{\sqrt{3}-1}\\
=\frac{2x(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\
=\frac{2x(\sqrt{3}+1)}{3-1}\\
=\frac{2x(\sqrt{3}+1)}{2}\\
=x(\sqrt{3}+1)\end{aligned}\end{array}$
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