In △ABC, AB=4, BC=5, and CA=6. In △EFG, EF=400, FG=500, and GE=600. Is △ABC similar to △EFG? If so, what is the scale factor
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Answer:
232°
Step-by-step explanation:
There are a couple of ways to find the desired direction. Perhaps the most straightforward is to add up the coordinates of the travel vectors.
30∠270° +50∠210° = 30(cos(270°), sin(270°)) +50(cos(210°), sin(210°))
= (0, -30) +(-43.301, -25) = (-43.301, -55)
Then the angle from port is ...
arctan(-55/-43.301) ≈ 231.79° . . . . . . . 3rd quadrant angle
The bearing of the ship from port is about 232°.
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<em>Comment on the problem statement</em>
The term "knot" is conventionally used to indicate a measure of speed (nautical mile per hour), not distance. It is derived from the use of a knotted rope to estimate speed. Knots on the rope were typically 47 ft 3 inches apart. As a measure of distance 30 knots is about 1417.5 feet.
