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Question

(C) Let $h = 30 \ m$ be the height of the lighthouse and $d$ be the distance of the boat from the foot of the lighthouse.In the right-angled triangle

Vedclass · Accepted Answer

$30 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right)$

Vedclass · Answer

(C) Let $h = 30 \ m$ be the height of the lighthouse and $d$ be the distance of the boat from the foot of the lighthouse.In the right-angled triangle formed,the angle of elevation of the top of the lighthouse from the boat is equal to the angle of depression,which is $15^\circ$.Thus,$	an 15^\circ = \frac{	ext{height}}{	ext{distance}} = \frac{30}{d}$.Therefore,$d = \frac{30}{	an 15^\circ}$.Since $	an 15^\circ = 2 - \sqrt{3}$,we have $d = \frac{30}{2 - \sqrt{3}}$.To rationalize the denominator,multiply the numerator and denominator by $(2 + \sqrt{3})$:$d = \frac{30(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{30(2 + \sqrt{3})}{4 - 3} = 30(2 + \sqrt{3})$.Alternatively,using $	an 15^\circ = 2 - \sqrt{3}$,we can also express the result as $d = \frac{30}{2 - \sqrt{3}} = 30(2 + \sqrt{3})$.Comparing with the given options,option $D$ is $30 \left( \frac{2 + \sqrt{3}}{2 - \sqrt{3}} ight) = 30(2 + \sqrt{3})^2$,which is not correct. However,checking the expression $d = \frac{30}{	an 15^\circ} = 30 \cot 15^\circ = 30(2 + \sqrt{3})$.Given the options,$30 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} ight) = 30 \frac{(\sqrt{3} + 1)^2}{3 - 1} = 30 \frac{4 + 2\sqrt{3}}{2} = 30(2 + \sqrt{3})$.Thus,the correct option is $C$.

From the top of a lighthouse $30 \ m$ high with its base at sea level,the angle of depression of a boat is $15^\circ$. The distance of the boat from the foot of the lighthouse is:

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