Question 813233: A plane leaves an airport and travels 2500 mi. due West, then turns 38 degrees to the south and travels 240 mi to a second airport. What is the displacement of the 1st airport from the 2nd airport? Round to 3 decimal places. It asks for Magnitude and Direction.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! use the law of cosines to find the third side of the triangle
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a^2 = b^2 + c^2 - 2bc cos(I)
where:
b = 2500
c = 240
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but first find the included angle I
I = 180 - 38 = 142 degrees
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a^2 = 2500^2 + 240^2 - 2*2500*240*cos(142)
a^2 = 6.25e6 + 5.76e4 - 1.2e6*(-0.7880107)
a^2 = 6.25e6 + 5.76e4 + 9.456129e5
a^2 = 7.2532129e6
a = 2693.179
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the airplane arrives at ap2 at a distance of 2693.179 miles from ap1.
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now use the law of sines to find the smaller angle J of the triangle (ap1, turn-point, ap2):
sin(J)/j = sin(I)/i
sin(J)/240 = sin(142)/2693.179
sin(J)= sin(142)240/2693.179
sin(J)= (0.6156614)240/2693.179
sin(J)= 0.054864
J = arcsin(0.054864)
J = 3.1450587 degrees
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direction from ap2 to ap1 is:
90 - 3.1450587 = 86.854941 ~= 86.855 degrees true north bearing
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Answer:
the vector from ap2 to ap1 is:
Direction: 86.855 degrees true north
Magnitude: 2693.179 miles
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