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12 votes
Two ships leave a harbor at

the same time. One ship travels
on a bearing S15°W at 10 miles per
hour. The other Ship travels on a
bearing N75°E at 10miles per hour.
bearing
How far apart will the ships be
after 3 hours?

User TommySM
by
5.6k points

1 Answer

8 votes

Answer:

30√3 ≈ 51.96 miles

Explanation:

The distance between the two ships can be found using the Law of Cosines, or using your knowledge of the side relationships in special triangles.

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Each ship is traveling at 10 mph, so after 3 hours will have traveled 30 miles.

The triangle OS1S2 formed by the harbor and the two ship locations is an isosceles triangle with base angles of 30°. Each half of OS1S2 is a 30-60-90 triangle whose longer leg is √3 times half the hypotenuse. The sum of those two "longer legs" is the distance between the ships.

The distance between ships is 2×15√3 = 30√3 ≈ 51.96 miles.

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Additional comment

If you prefer to use the Law of Cosines, you are looking for the length of the side opposite the 120° angle in a triangle with sides of 30 miles.

c² = 30² +30² -2·30·30·cos(120°) = 30²(2-2·(-0.5)) = 3·30²

c = 30√3 . . . . . take the square root (miles)

Two ships leave a harbor at the same time. One ship travels on a bearing S15°W at-example-1
User Santiago Gil
by
5.4k points