Let \(t=\tan\beta\), so \(\tan\alpha=3t\). By the tangent addition formula, \[\frac{3t+t}{1-3t^2}=2.\] Rearrangement gives \[3t^2+2t-1=0,\qquad(3t-1)(t+1)=0.\] If \(t=1/3\), the interval gives \(\beta\approx18.4^\circ\) and \(\alpha=45^\circ\). If \(t=-1\), it gives \(\beta=135^\circ\) and \(\alpha\approx108.4^\circ\). Thus the possible pairs are \[\boxed{(\alpha,\beta)=(45^\circ,18.4^\circ),\quad(108.4^\circ,135^\circ)}.\] Both pairs satisfy the original sum-angle equation.