[tex]v_r(t)=r\cdot i(t)=(0.5\,\Omega)(0.1\sin{(20t)\,A})=0.05\sin{(20t)}\,V\\\\v_l(t)=l\cdot \dfrac{d}{dt}(i(t))=(0.010\,H)(20\cdot 0.1\cos{(20t)\,A/s})=0.02\cos{(20t)\,V}[/tex]
Then the total voltage is ...
[tex]v(t)=(0.05\sin{(20t)}+0.02\cos{(20t)})\,V\approx 0.05385\sin{(20t+0.3805)\,V}[/tex]
The magnitude of the resultant voltage is √(.02²+.05²); the angle is arctan(.02/.05).