Oscilloscope trigger problem. the 'trigger' settings determine when the trace begins. suppose you have displayed a sine wave with v = vosin(? t)vosin(? t) on the screen, where vo=5vvo=5v and ? /2? =5khz? /2? =5khz. assume further that the signal triggers the scope when v=2.5vv=2.5v and ? t=30 degrees? t=30 degrees. example: suppose you double the amplitude of the wave without changing any oscilloscope settings. what are vv and ? t? t now when the scope is triggered? [ the level hasn't changed so the scope still triggers when v=2.5vv=2.5v. v=2.5=10sin(? t)v=2.5=10sin(? t); ? t=sin? 1(2.5/10)=14.4775121851511 degrees? t=sin? 1(2.5/10)=14.4775121851511 degrees.]question: assume that you reduce vovo back to 5 v and switch the trigger slope polarity from '+' to '-' without changing the trigger level dial. what are v and ? ? t when the scope is triggered now?