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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 A
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Solution(By Apex Team)

AB is a tower BD and BC are its shadows and CD = 2x The length of shadow height and distance . image $\begin{array}{l}\tan45^{\circ}=\frac{AB}{DB}\\ \Rightarrow1=\frac{h}{y}\Rightarrow y=h\\ \text{ and }\tan30^{\circ}=\frac{AB}{CB}\\ \begin{aligned}\Rightarrow\frac{1}{\sqrt{3}}=\frac{h}{2x+y}\\ \Rightarrow2x+y=\sqrt{3}h\\ \Rightarrow\sqrt{3}h-h=2x\\ \Rightarrow h(\sqrt{3}-1)=2x\\ \Rightarrow h=\frac{2x}{\sqrt{3}-1}\\ =\frac{2x(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\ =\frac{2x(\sqrt{3}+1)}{3-1}\\ =\frac{2x(\sqrt{3}+1)}{2}\\ =x(\sqrt{3}+1)\end{aligned}\end{array}$