a plane is flying at 290km/h [E 42° S] relative to the air when the wind velocity is 65km/h [E 25° N]. Calculate the velocity of the plane relative to the ground

320 km/h [E 31° S]

Explanation:

"E 42° S" means "east, 42° south", or 42° south from east.

"E 25° N" means "east, 25° north", or 25° north from east.

Taking east to be + x and north to be + y, the horizontal component of the velocity is:

vₓ = 290 cos (-42°) + 65 cos 25°

vₓ = 274 km/h

And the vertical component of the velocity is:

vᵧ = 290 sin (-42°) + 65 sin 25°

vᵧ = - 167 km/h

The magnitude is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (274) ² + (-167) ²

v = 320 km/h

And the direction is found with trig:

θ = atan (vᵧ / vₓ)

θ = atan (-167 / 274)

θ = - 31°

Therefore, the plane's velocity relative to the ground is 320 km/h [E 31° S].