Question : From the top of a building 60 metres high, the angles of depression of the top and bottom of a tower are observed to be $30^{\circ}$ and $60^{\circ}$, respectively. The height of the tower in metres is:
Option 1: 40
Option 2: 45
Option 3: 50
Option 4: 55
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Correct Answer: 40
Solution : Let $AB$ be the building and $CD$ be the tower such that $\angle BDE=30^{\circ},\angle BCA=60^{\circ}$ and $AB$ = 60 m. Let $CA = DE = x$ m. From $\Delta CAB$, $\frac{CA}{AB}=\cot 60^{\circ}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{x}{60}=\frac{1}{\sqrt{3}}$ $\Rightarrow x=60\times \frac{1}{\sqrt{3}}$ $\Rightarrow x=60\times \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=20\sqrt{3}$ m. $\Rightarrow CA=DE=20\sqrt{3}...(i)$ From $\Delta BED, $ $\frac{{BE}}{DE}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{BE}{20\sqrt{3}}=\frac{1}{\sqrt{3}} \quad [\text{using (i)}]$ $\Rightarrow BE=20\sqrt{3}\times \frac{1}{\sqrt{3}}$m = $20$ m $\therefore {CD=AE=AB-BE} =60$ m – $20$ m = $40$ m Hence, the correct answer is 40 m.
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