How to operate in Angle Notation (Electrical Engineering)

dumbdumNotSmart
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Background info (Not actual question): So I've been gone a week from university and I've missed a lot of content, one particular thing that boggles me is this new method to operate with vector magnitudes and their angles. The operations surprised me with the speed they were done. I did ask a classmate who'd been the whole week i missed, he couldn't muster the memory of how to operate either. So without further due the actual question:

QUESTION: We are operating in 2D space with phasors (Current,voltage,power etc.)
What are the general rules to add, subtract, multiply and divide phasors in angle notation (Degrees, not radians).

Equations a=|v|∠θ, b=|u|∠φ (all angles in degrees)
a+b=|?|∠?
 
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Remember how these things were done with complex numbers. If you have ##z_1 = a e^{i \theta_1}## and ##z_2 = a e^{i \theta_2}##, how would you do ##z_1 z_2##?
 
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Ah yes, convert to radians and we're in business
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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