Question

1. a cargo ship leaves port a on the east coast of the u.s. and travels on a course of 28° for 68 meters to get to port b. it then travels 95 meters on a course of 118° to reach port c. the ship will then head home to port a. find the distance from port c to port a and the total distance traveled.

Explanation:

Step1: Find the angle between the two - known sides

The first - course angle is $28^{\circ}$ and the second - course angle is $118^{\circ}$. The angle $\theta$ between the sides of lengths $AB = 68$ meters and $BC=95$ meters is $118^{\circ}-28^{\circ}=90^{\circ}$.

Step2: Use the Pythagorean theorem to find the distance from Port C to Port A

Let $AB = a = 68$, $BC = b = 95$. According to the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$, where $c$ is the distance from Port C to Port A.
$c=\sqrt{68^{2}+95^{2}}=\sqrt{4624 + 9025}=\sqrt{13649}\approx116.83$ meters.

Step3: Calculate the total distance traveled

The ship travels $68$ meters from Port A to Port B, $95$ meters from Port B to Port C, and then approximately $116.83$ meters from Port C to Port A.
The total distance $D=68 + 95+116.83=279.83$ meters.

Answer:

The distance from Port C to Port A is approximately $116.83$ meters and the total distance traveled is approximately $279.83$ meters.