Problem Statement
From the top of a cliff $150\text{ m}$ high, the angles of depression of two ships anchored in the sea in the same vertical plane and on the same side of the cliff are observed to be $30^\circ$ and $45^\circ$ respectively. Find the distance between the two ships (take $\sqrt{3} \approx 1.732$).
Verified Solution & Marking Scheme
Draw geometric diagram and identify angles of elevation
Let the cliff be $AB = 150\text{ m}$, where $A$ is the base on the sea level and $B$ is the top.
Let the closer ship be $C$ and the further ship be $D$ on the horizontal line $ACD$.
Angle of depression to ship $C = 45^\circ \implies \angle BCA = 45^\circ$.
Angle of depression to ship $D = 30^\circ \implies \angle BDA = 30^\circ$.
Calculate distance AC to the closer ship
In right $\triangle BAC$:
$\tan 45^\circ = \frac{AB}{AC} \implies 1 = \frac{150}{AC} \implies AC = 150\text{ m}$
Calculate distance AD to the further ship
In right $\triangle BAD$:
$\tan 30^\circ = \frac{AB}{AD} \implies \frac{1}{\sqrt{3}} = \frac{150}{AD} \implies AD = 150\sqrt{3}\text{ m}$
Find distance CD between the two ships
$CD = AD - AC = 150\sqrt{3} - 150 = 150(\sqrt{3} - 1)$
Substitute $\sqrt{3} = 1.732$:
$CD = 150(1.732 - 1) = 150 \times 0.732 = 109.8\text{ m}$