Question : The angle of depression of two ships from the top of a lighthouse is $60^{\circ}$ and $45^{\circ}$ towards the east. If the ships are 300 metres apart, the height of the lighthouse (in metres) is:
Option 1: $200\left ( 3+\sqrt{3} \right )$
Option 2: $250\left ( 3+{\sqrt3} \right )$
Option 3: $150\left (3 +{\sqrt3} \right)$
Option 4: $160 \left (3+{\sqrt3} \right)$
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Correct Answer: $150\left (3 +{\sqrt3} \right)$
Solution : In $\Delta ABC$ $\tan 60^{\circ} = \frac{AB}{BC}$ $⇒\sqrt{3} = \frac{AB}{BC}$ $⇒AB = \sqrt{3} BC \quad \text{-----------------------(1)}$ In $\Delta ABD$ $\tan 45^{\circ} = \frac{AB}{BD}$ $\Rightarrow 1 = \frac{AB}{BD}$ $⇒AB = BD \quad \text{---------------------------------(2)}$ From equation $(1)$ and equation $(2)$ $\Rightarrow BD = \sqrt{3} BC\quad \text{--------------------------(3)}$ From the figure $BD = BC + CD$ From equation $(3)$ $\Rightarrow \sqrt{3} BC = BC + 300$ $\Rightarrow BC(\sqrt{3} - 1) = 300$ $\Rightarrow BC = \frac{300}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$ $\Rightarrow BC = \frac{300}{2} \times (\sqrt{3} + 1)$ $\Rightarrow BC = 150 (\sqrt{3} + 1)$ From equation (1) $AB = 150 (\sqrt{3} + 1) \times \sqrt{3}$ $\therefore AB = 150 (3 + \sqrt{3})$ metres. Hence, the correct answer is $150 (3 + \sqrt{3})$ metres.
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