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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