Hello,
In the triangle ABC,
In the triangle ACB,
Two ships leave port at 10 A.M. One travels at a bearing of 50° at 25 mph, and the other travels at a bearing of 120° at 20 mph.
How far apart are the ships at noon? Round your answer to the nearest tenth.
1 answer:
The first ship travels (25 mi/h)*(2 h) = 50 mi.
The second ship travels (20 mi/n)*(2 h) = 40 mi.
The angle between their directions of travel is 120° -50° = 70°.
The law of cosines can be used to find the distance (d) between the ships.
d² = 50² + 40² - 2*50*40*cos(70°)
d² = 2500 +1600 -4000*cos(70°)
d² ≈ 2731.919
d ≈ √2731.919 ≈ 52.3 . . . . miles
At noon, the ships are about 52.3 miles apart.
The second ship travels (20 mi/n)*(2 h) = 40 mi.
The angle between their directions of travel is 120° -50° = 70°.
The law of cosines can be used to find the distance (d) between the ships.
d² = 50² + 40² - 2*50*40*cos(70°)
d² = 2500 +1600 -4000*cos(70°)
d² ≈ 2731.919
d ≈ √2731.919 ≈ 52.3 . . . . miles
At noon, the ships are about 52.3 miles apart.
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