Two forces, 474 N at 11deg and 425 N at 32deg. are applied to a car in an effort to accelarate. Magnitude = ?

Question

Two forces, 474 N at 11± and 425 N at 32± are
applied to a car in an effort to accelerate it.

m=3395 kg
Force 1= 474 Newtons at 11deg above horizontal.
Force 2= 425 Newtons at 32deg below horizontal.

1) What is the magnitude of the resultant of
these two forces? Answer in units of N.

2) Find the direction of the resultant force (in re-
lation to forward, with counterclockwise con-
sidered positive).
Answer in degrees from the positive x-axis,
with counter-clockwise positive, within the
limits of -180deg. to 180deg. Answer in units of deg.

3) If the car has a mass of 3395 kg, what accel-
eration does it have?
Ignore friction. Answer in units of m=s2.

Answer 1

In x-y coordinates
f1 = (465.291, 90.443),
f2 = (360.420, -225.216), a total of
ftot = (825.712, -134.772).
1),2): Magnitude and angle of ftot = 836.638 N, -9.270 deg.
3): Acceleration = 836.638/3395 = 0.246432 m/s^2.
I wonder how those forces are applied in real life. Maybe two guys pushing it.