Question

Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern lighthouse and N 21o E from the southern lighthouse, how much closer is the closest lighthouse to the boat than the lighthouse furthest away? The northern lighthouse is 51.3 miles closer than the southern lighthouse. The northern lighthouse is 24.4 miles closer than the southern lighthouse. The northern lighthouse is 79.9 miles closer than the southern lighthouse. The northern lighthouse is 17.0 miles closer than the southern lighthouse.

Answer 1

Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$, the southern lighthouse as $\rm S$, and the boat as $\rm B$. These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$), respectively. The three angles of any triangle add up to $180^{\circ}$. Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$.)

It is also given that the length between the two lighthouses (length of $\rm NS$) is $75\; \rm mi$.

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, $\angle {\rm B}$ is opposite to side $\rm NS$, whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$. Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$.

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$.

Rearrange and solve for the length of $\rm NB$:

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$.

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$, the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$.

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$.

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$.

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$, whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$. Hence the conclusion.