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From a ship at sea it is observed that the
angle subtended by feet $A$ and $B$ of two light houses, at the
ship is 30°. The ship sails 4 km towards $A$ and this angle
is then 48°, the distance of $B$ from the ship at the second
observation is

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6.460 km

6.472 km

6.476 km

6.478 km

answer is B.

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Detailed Solution

If $S_{1}$ and $S_{2}$ are the two positions of the ship
then $S_{1} S_{2} = 4$

$∠ A S_{1} B = 30^{\circ}$ and $∠ A S_{2} B = 48^{\circ}$

$\Rightarrow ∠ S_{1} B S_{2} = 18^{\circ}$ (Fig. 27.42)

Now from $Δ S_{1} B S_{2}$

$\begin{array}{l} \frac{S_{1} S_{2}}{\sin 18^{\circ}} = \frac{S_{2} B}{\sin 30^{\circ}} \\ \Rightarrow S_{2} B = \frac{4 \times (1 / 2)}{\sin 18^{\circ}} = 2 (\sqrt{5} + 1) = 6.472 . \end{array}$

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