An island is located 44 miles N33degrees23'W of a city. A freighter in distress radios its position as N10degrees42'E of the island and N15degrees13'W of the city. How far is the freighter from the city? Question content area bottom left Part 1 The freighter is enter your response here ▼ square miles miles cubic miles away from the city. (Round to the nearest whole number.) . . . Question content area right Part 1 33 degrees 23 ' 10 degrees 42 ' 15 degrees 13 ' N A C N

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°.