A jet plane is traveling with a velocity of 720 kilometers/hour due north. There is a crosswind of 40 kilometers/hour from the east. Find the resultant velocity and the direction of the jet plane. A.
720 kilometers/hour in the direction of 3.2° west of north
B.
720 kilometers/hour due north
C.
721 kilometers/hour in the direction of 3.2° west of north
D.
520 kilometers/hour in the direction of 45° east of north

Option c is correct v = 721 km/h θ = 3.2°

Explanation:

In pic is further explanation: A triangle means division

The magnitude of the velocity of the plane can be found by calculating the resultant of the two velocities:

 R=

The direction can be found as follows:  

0 = tan^-1 (720km ÷ h/40km ÷h) = 86.8

where 720 km/h represents the velocity towards north, while 40 km/h represents the velocity towards west, so the angle is measured with respect to east. Therefore:

the direction is 86.8*north of west

(Hope this helps can I pls have brainlist (crown)1

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