From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be \(30^{\circ}\) and \(60^{\circ}\). If the height of the lighthouse is \(z\) meters and the line joining the ships passes through the foot of the lighthouse, prove that the distance between the ships is \(\frac{4z}{\sqrt{3}} \ m\).
Answer:
Let \(CD\) denote the height of the lighthouse. \(A\) and \(B\) be the position of the two ships.
In \(\triangle CAD\), \(tan \ 30^{\circ} =\)
\(x =\) ---- (\(1\))
In \(\triangle CBD\), \(tan \ 60^{\circ} =\)
\(y =\) ---- (\(2\))
Distance between the ships \(=\)
\(=\) \(\frac{4z}{\sqrt{3}} \ m\).
Hence, proved.