Question : From the top of a cliff 100 metres high, the angles of depression of the top and bottom of a tower are 45° and 60°, respectively. The height of the tower is:
Option 1: $\frac{100}{3}(3-\sqrt{3})$ m
Option 2: $\frac{100}{3}(\sqrt3-1)$ m
Option 3: $\frac{100}{3}(2\sqrt3-1)$ m
Option 4: $\frac{100}{3}(\sqrt3-\sqrt{2})$ m
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Correct Answer: $\frac{100}{3}(3-\sqrt{3})$ m
Solution : Given: The top of a cliff 100 metres high, the angles of depression of the top and bottom of a tower are 45° and 60°. We know the trigonometric ratio, $tan\ \theta=\frac{\text{Height}}{\text{Base}}$. Let the height of the tower be $x$ m. In $\triangle AEC$, $\tan \ 45° = \frac{AE}{CE}$ $⇒1 = \frac{100 \:-\: x}{CE}$ $⇒CE=100-x$-------(equation 1) In $\triangle ABD$, CE = BD and $tan \ 60° = \frac{AB}{CE}$. $⇒\frac{\sqrt{3}}{1} = \frac{100}{(100 \:–\: x)}$ $⇒100-x = \frac{100}{\sqrt{3}}$ $⇒x=100(1-\frac{1}{\sqrt{3}})$ $\therefore x= \frac{100}{3}(3 - \sqrt{3})$ m Hence, the correct answer is $\frac{100}{3}(3-\sqrt{3})$ m.
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