A man of the deck of …

Let D be the position of the man

 Let AB be the deck of the ship above the water level and DE be the cliff
Let {tex}B E \perp C D{/tex} {tex}{/tex}
The angles of  {tex}{/tex}elevation of the top and depression of the base is 60^° and 30^°

CE = AB = 16m
Let CD = h meters
Then, ED = CD - CE

       = (h -16)m
From right {tex}\Delta B E D,{/tex} we have
{tex}\frac { B E } { E D } = \cot 60 ^ { \circ }{/tex}
{tex}\Rightarrow \frac { B E } { ( h - 16 ) } = \frac { 1 } { \sqrt { 3 } }{/tex}
{tex}\Rightarrow B E = \frac { ( h - 16 ) } { \sqrt { 3 } }{/tex}
From right {tex}\Delta \mathrm { CAB },{/tex}we have
{tex}\frac { A C } { A B } = \cot 30 ^ { \circ } \Rightarrow \frac { A C } { 16 } = \sqrt { 3 }{/tex}
{tex}\Rightarrow A C = 16 \sqrt { 3 } \mathrm { m }{/tex}
But BE = AC
{tex}\therefore{/tex} {tex}\frac { ( h - 16 ) } { \sqrt { 3 } } = 16 \sqrt { 3 } \Rightarrow ( h - 16 ) = 48{/tex}
{tex}\Rightarrow h = 64 m{/tex}
Hence the height of cliff is 64m and the distance between the cliff and the ship {tex}= 16 \sqrt { 3 } \mathrm { m } = 27.71 \mathrm { m }{/tex}