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I have two voltages given as v1(t) = 20cos([tex]\omega[/tex]t - 45)

and v2(t) = 10sin([tex]\omega[/tex]t + 60)

My task is to add them on the single form Vcos([tex]\omega[/tex]t + [tex]\theta[/tex])

The first part is relativley easy:

The phasors are v1 = 20[tex]\angle[/tex]-45)

and v2 = 10[tex]\angle[/tex]-30)

so i have 20[tex]\angle[/tex]-45) + 10[tex]\angle[/tex]-30)

20cos(-45) = 14.14

20sin(-45) = -j14.14

10cos(-30) = 8.66

10sin(-30) = -j5

Add those up and you have 22.8 -j19.14

But how do i get this in polar form?