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Home » As observed form the top of a lighthouse, above sea level, the angle of depression of a ship, sailing directly towards it, changes from and . Determine the distance travelled by the ship during the period of observation.
As observed form the top of a lighthouse, 100 \mathrm{~m} above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30^{\circ} and 60^{\circ}. Determine the distance travelled by the ship during the period of observation.
CBSE | Class 10 | Maths | RS Aggarwal | Some Applications Of Trigonometry
As observed form the top of a lighthouse, 100 \mathrm{~m} above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30^{\circ} and 60^{\circ}. Determine the distance travelled by the ship during the period of observation.

Let \mathrm{OA} be the lighthouse and \mathrm{B} and \mathrm{C} be the positions of the ship.

\mathrm{OA}=100 \mathrm{~m}, \angle \mathrm{OBA}=30^{\circ} and \angle \mathrm{OCA}=60^{\circ}

0 C=x m and B C=y m

In the right \triangle \mathrm{OAC}, we have

\begin{aligned} &\frac{O \mathrm{~A}}{\mathrm{OC}}=\tan 60^{\circ}=\sqrt{3} \\ &\Rightarrow \frac{100}{\mathrm{x}}=\sqrt{3} \\ &\Rightarrow \mathrm{x}=\frac{100}{\sqrt{3}} \mathrm{~m} \end{aligned}

in the right \triangle 0 B A, we have:

\begin{aligned} &\frac{0 \mathrm{~A}}{\mathrm{OB}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ &\Rightarrow \frac{100}{\mathrm{x}+\mathrm{y}}=\frac{1}{\sqrt{3}} \\ &\Rightarrow \mathrm{x}+\mathrm{y}=100 \sqrt{3} \end{aligned}

putting \mathrm{x}=\frac{100}{\sqrt{3}} in the above equation, we get:

y=100 \sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}=115.47 \mathrm{~m}

\therefore Distance travelled by the ship during the period of observation =\mathrm{B}=\mathrm{y}=115.47 \mathrm{~m}

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