Need help finding angular momentum of a particle

[Ella1777]

1. At the instant of the figure, a 6.70 kg particle P has a position vector

of magnitude 4.30 m and angle θ1 = 43.0° and a velocity vector

of magnitude 3.40 m/s and angle θ2 = 32.0°. Force

, of magnitude 7.40 N and angle θ3 = 32.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle 2. L=r x p=r x m x v3. φ3 = Θ1 + Θ3 = 43.0º + 32.0º = 75.0º
φ2 = 270º - Θ1 - Θ2 = 270º - 43º - 32º = 195º
φ1 = Θ1 = 43.0º


r = 4.30m*(cos43.0º i + sin43.0º j)
F = 7.40N*(cos75.0º i + sin75.0º j)
v = 3.40m/s*(cos195º i + sin195º j) L = r x p = r x m*v
where L, r, p and v are vectors.
L = 4.30m * 6.70kg * 3.40m/s * (cos43.0º i + sin43º j) x (cos195º i + sin195º j)
L = 4.30m * 6.70kg * 3.40m/s * (cos43.0*sin195 - sin43.0*cos195)k
L = L = 4.30m * 6.70kg * 3.40m/s * 0.469471562 k = 45.98661738 k kg·m²/s
which has magnitude 46.0 kg·m²/s
Although this seems to be wrong what am i doing wrong?

 

[kuruman]

The statement of the problem is hard to interpret. There are no images. Also, please use LaTeX for symbols.

 

[haruspex]

magnitude 3.40 m/s and angle θ2 = 32.0°

φ2 = 270º - Θ1 - Θ2 = 270º - 43º - 32

Since the images don't work, looks like you need to describe what this 32.0° angle is.
I would have guessed it meant the angle anticlockwise from the positive x axis, but that is not how you have treated it.

 

[Dr Dr news]

Where does the 195° come from? It looks like v = 3.40 m/s (cos 32° i + sin 32° j).

 

[haruspex]

Where does the 195° come from? It looks like v = 3.40 m/s (cos 32° i + sin 32° j).

Ella treated the 32 degrees as though it is additional to the 43 degrees, but without the diagram (or clear description) we cannot tell if that is appropriate.