SOLUTION: [ An airplane, which has airspeed of 575 km/hr, heads directly east. The wind is blowing with a velocity of 82.0 km/hr on a heading of 35.0 degrees West of South. What will the res

Question 819430: [ An airplane, which has airspeed of 575 km/hr, heads directly east. The wind is blowing with a velocity of 82.0 km/hr on a heading of 35.0 degrees West of South. What will the resulting velocity of this airplane be as measured from the ground? The answer is: 532 km/hr at 7.25 degrees SE ]
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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airplane vector a:
draw horizontal line (magnitude 575 kph) pointing to the right for east
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wind vector b:
at the tip of vector a draw vector b (magnitude 82 kph) at 55 degrees below a (90 - 35 = 55)
call this angle C
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the result vector c:
vectors a and b form two sides of a triangle, and the third side is vector c, the wind-corrected velocity of the airplane
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to find the magnitude of vector c use the law of cosines: c^2 = a^2 + b^2 - 2ab*cos(C)
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c^2 = a^2 + b^2 - 2ab*cos(C)
c^2 = 575^2 + 82^2 - 2(575)(82)*cos(55)
c = sqrt( 575^2 + 82^2 - 2(575)(82)*cos(55) )
c ~= 532.222 kph
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to find the angle B between vector a and vector c use the law of sines: a/sin(A) = b/sin(B) = c/sin(C)
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b/sin(B) = c/sin(C)
sin(B) = b*sin(C)/c
sin(B) = 82*sin(55)/532.222
B = arcsin( 82*sin(55)/532.222 )
B ~= 7.250 degrees
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answer:
using proper aviation bearing terminology, and avoiding antiquated references such as SE, West of South, etc., the wind-corrected velocity of the airplane relative to the ground, is:
532 kph at 97.3 degrees true north
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