Let AB be the lighthouse and the two ships be at point C and D respectively.
In ∆ABC,
\(\frac{AB}{ BC} = tan 45^{\degree}\)
\(\frac{75}{ BC} = 1\)
\(BC = 75\,m\)
In ∆ABD,
\(\frac{AB}{ BD}= tan 60^{\degree}\)
\(\frac{75}{ BC +CD} = \frac{1}{\sqrt3}\)
\(\frac{75}{ 75 + CD} = \frac1{ \sqrt3}\)
\(75 \sqrt3 = 75 + CD\)
\(75 (\sqrt3 -1)m = CD\)
Therefore, the distance between the two ships is \(75(\sqrt3 -1) \,m\).
The shadow of a tower on level ground is $30\ \text{m}$ longer when the sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower. (Use $\sqrt{3}=1.732$.)
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