Question

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 ${30}^0$, the ship sails $4km$ towards $A$' and this angle is then ${48}^0$, The distance of $B$' from the ship at the second observation is

• A. $8.460km$
• B. $6.472km$
• C. $8.476km$
• D. $7.478km$

Answer 1

Answer: (B)

$6.472km$

Since $S_1$ and $S_2$ are two position of ship then
$S_1{S}_2=4$
$\angle A{S}_{1}B={30}^{\circ },\angle A{S}_{2}B={45}^{\circ },\angle S_{1}B{S}_2={18}^{\circ }$
Now from $\Delta S_{1}B{S}_2$
$\frac{S_1{S}_2}{\sin {18}^{\circ }}=\frac{S_{2}B}{\sin {30}^{\circ }}$
$\Rightarrow S_{2}B=\frac{4\times \frac{1}{2}}{\sin {18}^{\circ }}$
$\Rightarrow S_{2}B=2(\sqrt{5}+1)=6.472$