If an ac voltage has a peak value of 155.6 V, what is the phase angle at which the instantaneous voltage is 110 V?
Write v=V_{M} \sin \theta (11-3)
Solve for θ: \sin \theta=\frac{v}{V_{M}} \\ \theta=\text{arcsin}\frac{v}{V_{M}}=\text{arcsin}\frac{100}{155.6}=0.707=45^{\circ}