Answer:
Let AB be the height of water level and CD be the height of hill Then, In \[\Delta \,ABC\] \[\Delta \,ABC\] \[\tan \,30{}^\circ =\frac{10}{y}\] \[y=10\sqrt{3}\] ?(i) In \[\Delta \,ADE\] \[\tan \,60{}^\circ =\frac{x}{y}\] \[y=\frac{x}{\sqrt{3}}\] ?(ii) From (i) and (ii), we get \[\frac{x}{\sqrt{3}}=10\sqrt{3}\] \[x=10\times 3=30\,m\] \[\therefore \] Distance of the hill from this ship is \[10\sqrt{3}\,m\] and the height of the hill is \[30+10=40\text{ }m\].
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