How to calculate the conduction angle of a CVD diode?
I'm writing this whole thing to supplement Spehro's answer. His answer is just fine. But I thought I'd add a picture, since you wrote, "I'm not sure where to start with the trig."



Hopefully, this circle rings a bell! For a sine wave voltage source, it is convenient to consider the angle \$0^{\circ}\$ in the same place that the mathematicians have used when defining the magnitude of the sine.

In the above case, your voltage source is the length of \$r\$. As the sine wave goes from \$0^{\circ}\$ to \$360^{\circ}\$, \$r\$ doesn't change at all. It's always the same size. The only thing we care about that does change, is the length of \$y\$. This is the momentary voltage from the voltage source. The equation is just:

$$y=r\cdot sin\left(\theta\right)$$

Okay. You also know, or you should, that when you are told the RMS voltage is \$X\$ then the peak voltage at the top of the sine wave must be \$\sqrt{2}X\$. That's also the magnitude for \$r\$. The only other thing you need to keep in your head is the voltage drop of the diode. That's given as \$700mV\$. So you know that when the voltage after the diode (across the resistor) just begins to exceed \$0V\$, that the value of \$y=r\cdot sin\left(\theta\right)\$ must be itself, right at that very same moment, just beginning to exceed \$700mV\$. So you set that up as an equation and follow through:

$$\begin{align*}
12V\sqrt{2}\cdot sin\left(\theta\right) &amp;= 700mV \\
sin\left(\theta\right) &amp;= \frac{700mV}{12V\sqrt{2}} \\
\theta &amp;= sin^{-1}\left(\frac{700mV}{12V\sqrt{2}}\right) \approx 2.364^{\circ}
\end{align*}$$

That's the angle at which the diode just starts to conduct. That's also the angle, just prior to \$180^{\circ}\$, that the diode will cease to conduct.

Hopefully, this reminds you about the trigonometry/geometry of the situation.