Question

two docks are located on an east-west line 2582 ft apart. from dock a, the bearing of a coral reef is 64°25. from dock b, the bearing of the coral reef is 334°25. find the distance from dock a to the coral reef.
the distance from dock a to the coral reef □ ft.
(round to the nearest integer as needed.)

Explanation:

Step1: Convert bearings to angles

First, convert the bearings to angles within the triangle formed by the two docks and the coral reef.

• From dock A, bearing $64^\circ25'$: The internal angle at A is $90^\circ - 64^\circ25' = 25^\circ35'$. Convert to decimal degrees: $25 + \frac{35}{60} \approx 25.5833^\circ$
• From dock B, bearing $334^\circ25'$: This is equivalent to $360^\circ - 334^\circ25' = 25^\circ35'$ east of north, so the internal angle at B is $90^\circ - 25^\circ35' = 64^\circ25'$. Convert to decimal degrees: $64 + \frac{25}{60} \approx 64.4167^\circ$

Step2: Find angle at the coral reef

Sum of angles in a triangle is $180^\circ$.
$\text{Angle at reef } C = 180^\circ - 25.5833^\circ - 64.4167^\circ = 90^\circ$

Step3: Use sine law to find distance

Let $d$ = distance from dock A to reef. The distance between docks is $2582$ ft.
By sine law: $\frac{d}{\sin(64.4167^\circ)} = \frac{2582}{\sin(90^\circ)}$
Since $\sin(90^\circ)=1$,
$d = 2582 \times \sin(64.4167^\circ)$
$\sin(64.4167^\circ) \approx \sin(64^\circ25') \approx 0.9025$
$d \approx 2582 \times 0.9025 \approx 2330$

Answer:

2330 ft