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Two sailboats leave a harbor in the Bahamas at the same time. The first sails at 20 mph in a direction 330°.?
The second sails at 34 mph in a direction 220°. Assuming that both boats maintain speed and heading, after 2 hours, how far apart are the boats?
I’m so bad at these questions, what formula can I use to figure this out?
make a triangle of the direction and the total distance covered.
you know one ang and length of the two arms forming that angle. so u the cos rule to solve the question.
here is the equation for cos rule
c^2 = a^2 + b^2 – 2abcox(theta)
c^2 = 68^2 + 40^2 – 2(68)(40)(cos110)
c = 89.91
the ships are 89.9 miles apart
3 Responses to “Two sailboats leave a harbor in the Bahamas at the same time. The first sails at 20 mph in a direction 330°.?”
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March 12th, 2010 at 11:04 am
cosine rule, draw the triangle
i think 89.9 miles apart
References :
March 12th, 2010 at 11:48 am
make a triangle of the direction and the total distance covered.
you know one ang and length of the two arms forming that angle. so u the cos rule to solve the question.
here is the equation for cos rule
c^2 = a^2 + b^2 – 2abcox(theta)
c^2 = 68^2 + 40^2 – 2(68)(40)(cos110)
c = 89.91
the ships are 89.9 miles apart
References :
March 12th, 2010 at 12:30 pm
The angle between these is 110. How I approached it is to check the angle between. Break it into components rcos< and rsin<. Since we want to make it simple put one vector at y=0 make that the 34. 34+cos20(.3420)20=40.84
20 (sin20).9397=18.794. 18.794^2 + 40.84^2=Resultant ^2
Multiply 2 hours.
It has been so long that I forgot how to use the cosine law.
References :