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Answer :
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.