Additive Subtraction

$\large \displaystyle | 1 + e^{j \theta} | = 0.8$

$\large \displaystyle |1 + \cos(\theta) + j \sin(\theta) | = 0.8$

$\large \displaystyle [1 + \cos(\theta)]^2 + [\sin^2(\theta)] = 0.64$

$\large \displaystyle \cos(\theta) = -0.68$

$\color{#3D99F6} \boxed{ \large \displaystyle \theta \approx 133^o }$

I suppose the starting angle is irrelevant, so we can use two phasors for the current with $e^{j\frac{\theta}{2}}$ and $e^{-j\frac{\theta}{2}}$ amps. (I picked R = 1 ohm for easy math without loss of generality.)

The currents sum together in the resistor, which gives a total current with magnitude $|e^{j\frac{\theta}{2}}+e^{-j\frac{\theta}{2}}|=2\cos{\frac{\theta}{2}}$ .

Setting this to equal to 0.8 amps and solving for $\theta$ gives $\theta = 2\arccos{\frac{0.8}{2}}=133^o$