Oscilloscope trigger homework problem

[Calcifer]

http://img502.imageshack.us/img502/8941/webworks4ez6.jpg" .

The first answer is 2.5V, but I'm having trouble getting ωt

I'm using the equation: V = V₀ sin(ωt) --> 2.5V = 5V sin(ωt) --> sin^-1(2.5V/5V) = ωt --> ωt = 30 degrees but that's incorrect and so is -30 degrees

I've just used an oscilloscope for the first time last week, and I don't know much about it. What does changing the trigger slope polarity affect in this case? Apparently the scope still triggers at 2.5V... please shed some light on me!

 

[Phrak]

A positive trigger at 2.5 volts means that scope should trigger on a rising part of the waveform as it reaches 2.5 volts. Negative trigger means it will trigger on a falling part.

It still triggers at 2.5 volts. But the waveform is falling.

 

[skeptic2]

Assuming Vo is equal to Vpeak then dividing the trigger voltage by Vpeak and finding its sine [ sin(Vt/Vo) ] will give you the phase angle.