A proton moving in a uniform magnetic field with v=1.18*10^6 m/s i hat experiences F=1.53*10^-16 N k hat

A second proton with v=2.3*10^6 j hat m/s experiences F= -3.69*10^-16 k hat N in the same field.

What is the magnitude of B?
What is the direction B as an angle measured couter clock wise from the +x axis?

Let Vector-B, having magnitude "B" T make an angle θ counterclockwise with i hat.
We have
(B*sin θ)*(1.18*10^6)*q = 1.53*10^-6, ------------------ 1 and
(B*cos θ)*(2.3*10^6)*q = -3.69*10^-6;--------------------- 2squaring and adding
(B^2)*(1.6^2)*(10^-38)(10^12)*{1.18^2 + 2.3^2} = (10^-12)*{1.53^2 +3.69^2} or
B^2 = 0.93*10^14 or B = 0.97*10^7 or 9.7*10^ T
Disregarding the sign dividing 1 by 2
θ = tan^-1[0.8081] = 38.94 or 38.9 degree