Avoiding a Tropical Storm---Law of Cosines

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greatwhiteshark

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A cruise ship keeps an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid a tropical storm, the captain heads out of San Juan in a direction of 20 degrees off a direction heading toward Barbados. The captian keeps the 15 knot speed for 10 hours, after which time the path to Barbados becomes clear of storms.

a) Through what angle should the captain turn to head directly to Barbados?

b) How long will it be before the ship reaches Barbados if the same 15 knot speed is maintained?
 
The law of cosines is
a² = b² + c² - 2bc*cos(A)
a is the last leg of the trip
A is the angle opposite that leg = 20°.
b = 15*10 = 150 miles
c = the direct route = 600 miles
Just plug in the numbers to get a. Then you can switch to the law of sines
sin(A)/a = sin(C)/c or stick with cosines (changing letters and solving for cos(C)) with
cos(C) = (c² - a² - b²)/(2ab)

a.) C is the angle of the triangle so he turns (180 - C)° to get there.

b.) It will take 10 + c/15 hours for the total trip.
 
Re: Tell me...

greatwhiteshark said:
There are three different law of cosines.
Click to expand...
There is only one. Maybe this version.

a = Known length of one side of a triangle
b = Known length of another side of the same triangle
c = Unknown length of the last side of the same triangle.

w = angle formed by 'a' and 'b' = angle opposite side 'c'

c^2 = a^2 + b^2 - 2*a*b*cos(w)

Perhaps your confusion is in defining 'a', 'b', 'c' and 'w' before you know what you need.

Stuff we know
c^2 = a^2 + b^2 - 2*a*b*cos(w)

Stuff we don't know
c^2 = a^2 + b^2 - 2*a*b*cos(w)
 
Yes...

This is exactly what I was looking for. I never ask for my questions to be solved. I want to learn how to solve my homework questions alone. I just need help setting up certain questions occassionally.
 
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