Two ships leave a harbor at
1 answer:
Answer:
30√3 ≈ 51.96 miles
Step-by-step 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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<em>Additional comment</em>
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)
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