Question

From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be 60^o .If the height of the lighthouse is 90 m then find how far is that ship from the lighthouse?
($\sqrt{3}=1.73$)

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{ship}\mathrm{from}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{light}\mathrm{house}\mathrm{be}x\mathrm{m}. \\ \mathrm{Given}:\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{light}\mathrm{house},PQ=90\mathrm{m}\mathrm{and}\mathrm{angle}\mathrm{of}\mathrm{depression},\angle R=\theta =60° \\ \mathrm{We}\mathrm{know}\mathrm{that}\tan \theta =\frac{\mathrm{perpendicular}}{\mathrm{base}}=\frac{PQ}{QR} \\ \Rightarrow \tan 60°=\frac{90}{x} \\ \Rightarrow \sqrt{3}=\frac{90}{x} \\ \Rightarrow x=\frac{90}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \Rightarrow \mathrm{x}=\frac{90\sqrt{3}}{3}=30\sqrt{3} \\ \Rightarrow \mathrm{x}=30\times 1.73=51.9 \\ \therefore \mathrm{The}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{ship}\mathrm{from}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{light}\mathrm{house}\mathrm{is}51.9\mathrm{m}. \end{gathered}$$