Question

A cruise ship leaves its port at a heading of 135°. It travels for 400 miles, then turns to a heading of 180°. After another 250 miles, it reaches an Island N ¢ port W+ E 400 miles 21 1359 250 miles island S What is the approximate distance between the port and the Island?

Answer 1

Answer:

603.26 miles.

Explanation:

By the cosine law, we can find the distance from the port to the island as follows

$c^2=a^2+b^2-2ab\cos C$

Where c is the distance from the port to the island, a and b are the other sides of the triangle with lengths 400 mi and 250 mi and C is the angle between them of 135 degrees.

So, replacing the values, we get:

$\begin{gathered} c^2=400^2+250^2-2(400)(250)\cos 135_{} \\ c^2=160000+62500-(200000)(-0.7071) \\ c^2=160000+62500+141421.35 \\ c^2=363921.35 \\ c=\sqrt[]{363921.35} \\ c=603.26\text{ mi} \end{gathered}$

Therefore, the approximate distance between the port and the island is 603.26 miles.