Answer:
You get the SAME answer from both equations:
[tex]|V_A|=\sqrt{19} \,\,m/s\approx 4.36\,\,m/s[/tex]
Explanation:
Notice that you are dealing with vector equations, and that the first one doesn't give you 5 m/s for the magnitude.
You need to add V0 = 3 (in the x-direction), to
[tex]V_{AO}=2\,cos(60^o)\,\hat i+ 2\,sin(60^o)\,\hat j\\V_{AO}=2\,\frac{1}{2} cos(60^o)\,\hati + 2\,\frac{\sqrt{3} }{2} \,\hat j\\V_{AO}=1\,\,\hat i + 2\,\sqrt{3} \,\hat j[/tex]
Therefore, the answer for VA from the first equation is:
[tex]V_A=V_0\,+\,V_{AO}\\V_A=4\,\,\hat i + 2\,\sqrt{3} \,\hat j\\|V_A|=\sqrt{16+3} =\sqrt{19} \approx 4.36\,\,m/s\\\theta = arctan(\frac{\sqrt{3} }{4} )\approx 23.41^o[/tex]
which is exactly what was calculated in the second approach considering from point C.
