Magnetic field PHYSICS HELP PLEASE!

Question

A 0.40 uC particle moves with a speed of 16 m/s  through a region where the magnetic field has a strength of 0.99 T.

a.   At what angle to the field is the particle moving if the force exerted on it is 4.8x10^-6 N?

b. At what angle to the field is the particle moving if the force exerted on it is 3.0x10^^-6 N?

c. At what angle to the field is the particle moving if the force exerted on it is 1.0x10^-7N ?

Answer 1

the lLorentz force, is F=q*vXB

F=4.8x10^-6 N B=0.99

V=16m/s q=0.4uC what is angle

4.8*10^6 = 0.4*10^-6*16*Bsin(x)

X=sin^-1(x)=F/q*v

get 49 deg

please let me know if you agree