Question

two docks are located on an east - west line 2580 ft apart. from dock a, the bearing of a coral reef is 65°23. from dock b, the bearing of the coral reef is 335°23. find the distance from dock a to the coral reef. (round to the nearest integer as needed.)

Explanation:

Step1: Convert angles to decimal degrees

First, convert the angles from degrees - minutes to decimal degrees.
$65^{\circ}23' = 65+\frac{23}{60}\approx65.3833^{\circ}$
$335^{\circ}23'=335+\frac{23}{60}\approx335.3833^{\circ}$. The angle between the two - lines of sight from the docks to the reef is $\theta = 335.3833 - 65.3833=270^{\circ}$. Let the distance between the two docks be $AB = 2580$ ft. Let the distance from dock A to the reef be $x$, from dock B to the reef be $y$, and the angle at dock A be $\alpha=65.3833^{\circ}$, the angle at dock B be $\beta = 180 - 335.3833+360=104.6167^{\circ}$ (the interior - angle at B). The third angle of the triangle formed by the two docks and the reef is $\gamma=180-(65.3833 + 104.6167)=10^{\circ}$.

Step2: Use the Law of Sines

By the Law of Sines, $\frac{x}{\sin\beta}=\frac{AB}{\sin\gamma}$.
We know that $AB = 2580$ ft, $\beta = 104.6167^{\circ}$, and $\gamma = 10^{\circ}$.
So $x=\frac{AB\sin\beta}{\sin\gamma}=\frac{2580\times\sin(104.6167^{\circ})}{\sin(10^{\circ})}$.
Since $\sin(104.6167^{\circ})\approx0.97$, $\sin(10^{\circ})\approx0.174$.
$x=\frac{2580\times0.97}{0.174}$.
$x=\frac{2502.6}{0.174}=14497.701\approx14498$ ft.

Answer:

14498