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As observed from the top of a $70\text{ m}$ high lighthouse from the sea level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same sides of the lighthouse, find the distance between the two ships. (Use $\sqrt{3} = 1.73$)
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Correct figure [$1$ mark]
Let AB be the light house & $S_1$ and $S_2$ be the position of the two ships.
In $\Delta BAS_1$, $\tan 45^\circ = \frac{70}{y} = 1 \Rightarrow y = 70\text{ m}$ [$1\frac{1}{2}$ mark]
In $\Delta BAS_2$, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{70}{y + x} \Rightarrow x + y = 70\sqrt{3}$ [$1\frac{1}{2}$ mark]
$\Rightarrow x = 70 (1.73 - 1) = 51.1\text{ m}$ [$1$ mark]
Hence the distance between the the two ships is $51.1\text{ m}$.
Let AB be the light house & $S_1$ and $S_2$ be the position of the two ships.
In $\Delta BAS_1$, $\tan 45^\circ = \frac{70}{y} = 1 \Rightarrow y = 70\text{ m}$ [$1\frac{1}{2}$ mark]
In $\Delta BAS_2$, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{70}{y + x} \Rightarrow x + y = 70\sqrt{3}$ [$1\frac{1}{2}$ mark]
$\Rightarrow x = 70 (1.73 - 1) = 51.1\text{ m}$ [$1$ mark]
Hence the distance between the the two ships is $51.1\text{ m}$.