Answer:
70 miles
Step-by-step explanation:
Given an island is 44 miles N33°23'W of a city, and a freighter is N10°42'E of the island and N15°13'W of the city, you want to know the distance of the freighter from the city.
Triangle
As in the attachment, the locations of the city (C), island (I), and freighter (F) comprise the vertices of a triangle. The interior angles of this triangle are ...
C = N33°23'W - N15°13'W = 18°10'
F = N15°13'W -N10°42'E = 15°13' +10°42' = 25°55'
I = 180° -C -F = 180° -45°5' = 135°55'
Side CI of the triangle is given as 44 miles.
Law of sines
The length CF of the triangle can be found from the law of sines:
[tex]\dfrac{CF}{\sin(I)}=\dfrac{CI}{\sin(F)}\\\\\\CF=CI\cdot\dfrac{\sin(I)}{\sin(F)}=44\cdot\dfrac{\sin(135^\circ55')}{\sin(25^\circ55')}\approx 70.038[/tex]
The freighter is about 70 miles away from the city.
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Additional comment
If your calculator doesn't work with degrees and minutes, you can use the fact that a minute is 1/60 of a degree. Then 55' = 55/60° ≈ 0.91667°.
