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# As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are 30º and 45º. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Let AB be the lighthouse of height 75 m and let two ships be at C and D such that the angles of depression from B are 45º and 30º respectively.

Let AC = x and CD = y.

In $$∆ \ ABC$$, we have

$$\frac{AB}{AC} \ = \ tan45º \$$
$$=> \ \frac{75}{x} \ = \ 1$$
$$=> \ x \ = \ 75$$ (1)

In $$∆ \ ABD$$, we have

$$\frac{AB}{AD} \ = \ tan30º \$$
$$=> \ \frac{75}{x+y} \ = \ \frac{1}{ \sqrt{3}}$$
$$=> \ x+y \ = \ 75 \sqrt{3}$$ (2)

From (1) and (2), we have

$$75+y \ = \ 75 \sqrt{3} \$$
$$=> \ y \ = \ 75( \sqrt{3} - 1) \$$ T
$$=> \ y \ = \ 75(1.732 - 1) \ = \ 54.9$$

$$\therefore$$ the distance between the two ships is 54.9 metres.