Introduction to Trigonometry and Its Applications Question - CBSE

Question

As observed from the top of a $75	ext{ m}$ high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\sqrt{3} \approx 1.732$)

Accepted Answer

Let lighthouse be $AB = 75 ext{ m}$. Ships are at $C$ (farther, angle of elevation $= 30^\circ$) and $D$ (nearer, angle of elevation $= 45^\circ$). [1.0 Mark]
In right $\Delta ABD$: $ an 45^\circ = \dfrac{AB}{BD} \Rightarrow 1 = \dfrac{75}{BD} \Rightarrow BD = 75 ext{ m}$. [1.0 Mark]
In right $\Delta ABC$: $ an 30^\circ = \dfrac{AB}{BC} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{75}{BC} \Rightarrow BC = 75\sqrt{3} ext{ m}$. [1.5 Marks]
Distance between ships $CD = BC - BD = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) ext{ m}$. [1.0 Mark]
$= 75(1.732 - 1) = 75 imes 0.732 = 54.9 ext{ m}$. Distance between ships is $54.9 ext{ m}$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Diagrammatic triangle setup and angle identification: 1.0 Mark
Finding nearer ship distance $BD = 75 ext{ m}$: 1.0 Mark
Finding farther ship distance $BC = 75\sqrt{3} ext{ m}$: 1.5 Marks
Calculating distance between ships $75(\sqrt{3}-1)$: 1.0 Mark
Evaluating numerical value $= 54.9 ext{ m}$: 0.5 Mark

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