From the top of a 60 m high lighthouse with | vedclass

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(B) Let $AC$ be the lighthouse of height $60 \ m$ and $B$ be the position of the boat.Let $x$ be the distance of the boat from the base of the lightho

Vedclass · Accepted Answer

$60\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) \ m$

Vedclass · Answer

(B) Let $AC$ be the lighthouse of height $60 \ m$ and $B$ be the position of the boat.Let $x$ be the distance of the boat from the base of the lighthouse $(AB = x)$.The angle of depression from the top $C$ to the boat $B$ is $15^{\circ}$,so the angle of elevation from the boat to the top is also $15^{\circ}$.In $	riangle ABC$,$	an 15^{\circ} = \frac{AC}{AB} = \frac{60}{x}$.Therefore,$x = \frac{60}{	an 15^{\circ}}$.We know that $	an 15^{\circ} = 	an(45^{\circ} - 30^{\circ}) = \frac{	an 45^{\circ} - 	an 30^{\circ}}{1 + 	an 45^{\circ} 	an 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.Substituting this into the equation for $x$:$x = 60 \div \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}ight) = 60 \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}ight) \ m$.

From the top of a $60 \ m$ high lighthouse with its base at sea level,the angle of depression of a boat is $15^{\circ}$. The distance of the boat from the lighthouse is

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