The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is.
A. $100 \mathrm{~m}$ B. $100 \sqrt{3} m$ C. $100(\sqrt{3}-1) m$ D. $\frac{100}{\sqrt{3}} m$ Answer: Option CShow Answer
Solution(By Apex Team)
Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y
$\begin{array}{l}\text{Now in right }\ \triangle\text{ADB},\\
\tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{DB}\\
\tan45^{\circ}=\frac{100}{y}\\
\Rightarrow1=\frac{100}{y}\\
\Rightarrow y=100\\
\text{Similarily in right }\triangle\text{ACB},\\
\tan 30^{\circ}=\frac{A B}{C B} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{y+x} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{100+x} \\
\Rightarrow 100+x=100 \sqrt{3} \\
\Rightarrow x=100 \sqrt{3}-100 \\
\Rightarrow x=100(\sqrt{3}-1) m
\end{array}$
$\begin{array}{l}\text{Now in right }\ \triangle\text{ADB},\\
\tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{DB}\\
\tan45^{\circ}=\frac{100}{y}\\
\Rightarrow1=\frac{100}{y}\\
\Rightarrow y=100\\
\text{Similarily in right }\triangle\text{ACB},\\
\tan 30^{\circ}=\frac{A B}{C B} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{y+x} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{100+x} \\
\Rightarrow 100+x=100 \sqrt{3} \\
\Rightarrow x=100 \sqrt{3}-100 \\
\Rightarrow x=100(\sqrt{3}-1) m
\end{array}$
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