Discussion

Explanation:

Let LM be the lighthouse and B1 and B2 be the positions of the two boats.

In ∆LMB1,
Tan 30° = LM/MB1
⇒ MB1 = LM√3 = 300√3

Also, in ∆LMB2,
Tan 45° = LM/MB2
⇒ MB2 = LM = 300

In ∆MB1B2
(B1B2)2 = (MB1)2 + (MB2)2
⇒ (B1B2)2 = (300√3)2 + (300)2
⇒ (B1B2)2 = 3002 × [(√3)2 + (1)2]
⇒ (B1B2)2 = 3002 × 4
⇒ B1B2 = 300 × 2 = 600

Hence, option (d).

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