A plane is traveling 45 degrees N of E at 320 km/h when it comes across a current from S of E at 115 degrees of 20 km/h. What are the airplane's new course and speed?

Correct answer:

Φ =  48.291 °
v =  327.3803 km/h

Step-by-step explanation:

α=45  v1=320 km/h β=115  v2=20 km/h  n=v1 cosα+v2 cosβ=v1 cos45° +v2 cos115° =320 cos45° +20 cos115° =320 0.707107+20 (0.422618)=217.8218 e=v1 sinα+v2 sinβ=v1 sin45° +v2 sin115° =320 sin45° +20 sin115° =320 0.707107+20 0.906308=244.40033  tan Φ = e : n  Φ=π180°arctan(e/n)=π180°arctan(244.4003/217.8218)=48.291=48°1728"
v2 = n2+e2 v=n2+e2=217.82182+244.40032=327.3803 km/h

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