Asked by Adam
An airplane at 410 miles per hour at 40 degrees east of south is blown off course by a wind of 50 miles per hour at 25 degrees east of south. Find the resultant speed of the airplane.
Answers
Answered by
Henry
Vp + Vw = 410mi/h[310o]+50mi/h[295o].
X = 410*cos310 + 50*cos295 = 284.7 mi/h
Y = 410*sin310 + 50*sin295 = -359.4 mi/h
tanA = Y/X = -359.4/284.7 = -1.26236
A = -51.6 = 51.6o S. of E.
Vr = Y/sin(-51.6) = 459 mi/h
X = 410*cos310 + 50*cos295 = 284.7 mi/h
Y = 410*sin310 + 50*sin295 = -359.4 mi/h
tanA = Y/X = -359.4/284.7 = -1.26236
A = -51.6 = 51.6o S. of E.
Vr = Y/sin(-51.6) = 459 mi/h
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