The calculated angle is accidental, it is not a reliable value. To demonstrate how unreliable is it, imagine that capacitor is 1 pF. The angle will be close to zero, << much less << than 270.
I'd suggest to add optocoupler with high voltage diode and resistor, read zero crossing with accurate estimate of current transfer ratio of optocoupler, known part's delays, and known software delays. Though with high voltages (above 300V) the coupler should be super low current kind, say with LED requiring less than 1 mA to be detectable.
Yes, use a 12VDC input (>6.5A or more), then regulate it down to 5V for the low voltage section.
You can use a simple fixed linear regulator like an LM7085 for this, or for better efficiency use a buck regulator like the TL2575 (lots more options here).
EDIT - more detail on the difference between linear and switching regulator options.
With a linear regulator, it drops and regulates the voltage by simply dissipating the difference as heat.
For example in your case of 12V in and 5V out at 1A, the dissipation will be:
(12 - 5) * 1A = 7W (as Steven also calculated in his answer)
So do we need a heatsink? how hot will the regulator get?
In the datasheet for the LM7805, the thermal resistance for junction to air (Rθja) is given as 65°C/W. This means that for every watt dissipated, the junction temperature will rise by 65°C above ambient.
So at 7W, we get 7 * 65°C = 455°C = way too hot! (the regulator will actually shut down at ~150°C to protect itself)
The absolute maximum temperature before damage is given as 150°C, and maximum operating temperature is given as 125°C, so you would need a reasonably sized heatsink to keep the temperature within limits.
To calculate the heatsink needed, you take the junction to case rating (Rθjc), add this to the case to heatsink rating (Rθc-hs) and heatsink to air rating (Rθhs).
Here is an introduction to heatsink selection:
Heatsink Basics
A switching regulator is a different story. Switching regulators regulate by transforming the power rather than dissipating it (much like a transformer)
If we ignore the small inefficiencies for a moment and assume 100%, we know that power in must equal power out.
So if voltage in = 12V, and voltage out equals 5V and needs to be 1A, we know that 5W of power is needed at the output.
With this information we can calculate the current in:
5W / 12V = 417mA.
Now lets add the inefficiency (say 88%, from the TL2575 datasheet linked to above) and calculate:
5W / 0.88 = 5.68W at the input
5.88W / 12V = 473mA input current
5.68W - 5W = 680mW dissipated as heat.
How about the heatsink?
The TL2575 has a thermal resistance of 31.8°C/W, so:
31.8 * 0.68 = 21.62°C rise above ambient.
If the ambient is 20°C, the temperature will be 41.62°C.
Even at 50°C ambient, we would still be well within operating limits.
So you can see using the switching regulator in this case makes sense. You can go for a "ready rolled" module like in Steven's answer, or build your own using something like the TL2575 IC linked to above. The datasheet will have a few example circuits to help you along (note that a "buck regulator" is a type of switching regulator that lowers the voltage. A "boost regulator" is a switching regulator that raises the voltage)
Best Answer
Actually, isolation is not a must for UL certification. So yes, a transformerless regulator can be certified. You can find a lot of non-isolated UL-listed products on the market.
Probably the reviewer has no knowledge about capacitive power supplies but anyways it seems that s/he treats C2 as a "Safety-Critical Component" due to its position (i.e. mains side - after the bridge rectifier). S/he may also suggest you to use even an X/Y cap there.
In one (or some) of the EN-.... standards (I cannot really remember, sorry) safety-critical components should be defined as:
etc.
From my point of view, the safety-critical components are R1, R2 and C1 (and a varistor, if any) in your circuit even if it has a zener at its output.
I think it's hard to find a standard document (because nearly all of them are protected against public access) but you can persuade the reviewer about the voltage rating about C2 by showing some measurements, test results, technical documents etc.