Answer
$i_3 = 143.2cos(377t+2)$ mA
Work Step by Step
This problem is is intuitive given you have dealt with KCL and KVL.
We know that the total current coming into a node and leaving a node must equal 0.
$\sum I = i_1 - i_2 - i_3 = 0$
solve for $i_3$
$i_3 = i_1 - i_2$
$i_3 = 141.4cos(377t + 2.356) - 50sin(377t - 0.927)$
Convert to phasor notation after converting the sin to a cos with a phase shift
$I_3(377t) = 141.4e^{2.356j} + 50e^{0.6438j} $
Use a calculator with complex num capabilities to calculate the final result
$i_3 = 143.2cos(377t + 2) $mA
