Solution using Euler

-> 500cos(a)-288.6sin(a)-300=0

Euler:
    cos(x) = (e^jx+e^-jx)/2
    sin(x) = (e^jx-e^-jx)/2j

-> Substitute in the original equation:

250e^ja+250e^-ja-(144.3/j)e^ja+(144.3/j)e^-ja-300=0

-> Reorder terms:

(250-144.3/j)e^ja+(250+144.3/j)e^-ja-300=0

-> Multiply by e^ja:

(250-144.3/j)e^2ja-300e^ja+(250+144.3/j)=0

-> Now solve the equation to find e^ja

a=((-144.3+250j)/j)=250+144.3j
b=-300
c=((144.3+250j)/j)=250-144.3j

sol=(-b+-sqrt(b^2-4ac))/2a

-> We have two solutions:

ans1=0.87716506966622945779+0.48018896338570592133j
ans2=0.02295205357384654090-0.99973656691987778778j

For ans1:
----------
e^ja=ans1
ja=ln(ans1)
  =ln(0.87716506966622945779+0.48018896338570592133j)
  =0.50087012481978262042j
a=0.50087012481978262042 rad (28.6977442 degrees)

For ans2:
----------
e^ja=ans2
ja=ln(ans2)
  =ln(0.02295205357384654090-0.99973656691987778778j)
  =-1.54784225756526295179j
a=-1.54784225756526295179 (-88.6848287 degrees)