Solution :
Let $\phi$ be the phase difference betweeen applied voltage and the current
$\tan \phi = \large\frac{X_L- X_C}{R}$
$\qquad= I_v \large\frac{(X_L-X_C)}{I_vR}$
$\qquad= \large\frac{V_L -V_C}{V_R}$
$\qquad= \large\frac{30-30}{60}$
$\tan \phi = zero$
$\phi = 0^{\circ}$