The shadow of a tower standing on a level ground is found to be $60 \mathrm{~m}$ longer when the Sun's…

The shadow of a tower standing on a level ground is found to be $60 \mathrm{~m}$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is
  1. $30 \mathrm{~m}$
  2. $90 \mathrm{~m}$
  3. $60 \sqrt{3} \mathrm{~m}$
  4. $30(\sqrt{3}+1) \mathrm{m}$

Solution

In $\triangle B A C$,
$\begin{array}{rlrl} & & \tan 30^{\circ} & =\frac{B A}{60+x}=\frac{h}{60+x} \\ \Rightarrow & \frac{1}{\sqrt{3}} & =\frac{h}{60+x} \\ \Rightarrow & & 60+x & =\sqrt{3} h \\ \text { Also, } & & \tan 45^{\circ} & =\frac{h}{x} \Rightarrow h=x\end{array}$ From Eqs. (i) and (ii), we get $\begin{aligned} & \Rightarrow & 60+h & =\sqrt{3} h \\ & \Rightarrow & h & =\frac{60}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ & \therefore & h & =30(\sqrt{3}+1)\end{aligned}$

Asked in: AP EAMCET 2001

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