Picture the mass moving up and down under the harmonic force \(F_0 \sin(\omega t)\). The displacement response can be written as a wave that lags the force by some angle \(\phi\), where \(\tan \phi = \dfrac{2 \zeta r}{1 - r^2}\) and \(r = \omega/\omega_n\) is the frequency ratio. Resonance happens exactly when the driving frequency matches the natural frequency, so \(r = 1\). Putting \(r = 1\) into the formula makes the denominator \(1 - r^2\) go to zero, so the tangent of the phase angle shoots up to infinity. The only angle in this range where \(\tan \phi\) is infinite is \(90^\circ\). This also matches the physical picture: at resonance, the applied force is used up entirely in beating the damping force, and since the damping force is always in phase with velocity, the displacement ends up a quarter cycle, that is \(90^\circ\), behind the force. Hence the answer is \(90^\circ\), option (3).