QuestionA ship is sailing east. At one point, the bearing of a submerged rock is $46^{\circ} 20^{\prime}$. After the ship has sailed 13.7 mi , the bearing of the rock has become $307^{\circ} 40^{\prime}$. Find the distance of the ship from the rock at the latter point.
The distance is approximately $\square$ mi. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Studdy Solution
Use the Law of Cosines to find the distance from the ship to the rock:
The Law of Cosines states:
$$c^2 = a^2 + b^2 - 2ab \cos(C)$$
Where: - $a = 13.7$ miles (distance sailed), - $b$ is the distance from the ship to the rock at the latter point, - $C$ is the angle between the two bearings.
Calculate the angle $C$:
$$C = 360^\circ - (43.667^\circ + 52.333^\circ) = 264^\circ$$
Substitute into the Law of Cosines:
$$b^2 = 13.7^2 + 0^2 - 2 \times 13.7 \times 0 \times \cos(264^\circ)$$
Since the second term becomes zero due to the distance being zero initially, we simplify:
$$b^2 = 13.7^2$$
$$b = \sqrt{13.7^2}$$
$$b = 13.7 \text{ mi}$$
The distance of the ship from the rock at the latter point is:
$$\boxed{13.7 \text{ mi}}$$