Answer:
The northern lighthouse is approximately [tex]24.4\; \rm mi[/tex] closer to the boat than the southern lighthouse.
Step-by-step explanation:
Refer to the diagram attached. Denote the northern lighthouse as [tex]\rm N[/tex], the southern lighthouse as [tex]\rm S[/tex], and the boat as [tex]\rm B[/tex]. These three points would form a triangle.
It is given that two of the angles of this triangle measure [tex]40^{\circ}[/tex] (northern lighthouse, [tex]\angle {\rm N}[/tex]) and [tex]21^{\circ}[/tex] (southern lighthouse [tex]\angle {\rm S}[/tex]), respectively. The three angles of any triangle add up to [tex]180^{\circ}[/tex]. Therefore, the third angle of this triangle would measure [tex]180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}[/tex] (boat [tex]\angle {\rm B}[/tex].)
It is also given that the length between the two lighthouses (length of [tex]\rm NS[/tex]) is [tex]75\; \rm mi[/tex].
By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, [tex]\angle {\rm B}[/tex] is opposite to side [tex]\rm NS[/tex], whereas [tex]\angle {\rm S}[/tex] is opposite to side [tex]{\rm NB}[/tex]. Therefore:
[tex]\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}[/tex].
Substitute in the known measurements:
[tex]\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}[/tex].
Rearrange and solve for the length of [tex]\rm NB[/tex]:
[tex]\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}[/tex].
(Round to at least one more decimal places than the values in the choices.)
Likewise, with [tex]\angle {\rm N}[/tex] is opposite to side [tex]{\rm SB}[/tex], the following would also hold:
[tex]\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}[/tex].
[tex]\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}[/tex].
[tex]\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}[/tex].
In other words, the distance between the northern lighthouse and the boat is approximately [tex]30.73\; \rm mi[/tex], whereas the distance between the southern lighthouse and the boat is approximately [tex]55.12\; \rm mi[/tex]. Hence the conclusion.
