Answer:
[tex]|A|=32.2048438m[/tex]
[tex]\alpha =25.76932762$^{\circ}$[/tex]
Explanation:
In order to find the magnitude of the vector we need to use pythagorean theorem. As you can see in the picture I attached you, the components Ax and Ay represents the legs and the the hypotenuse is the magnitude of the vector A. Hence:
[tex]|A|=\sqrt{(Ax^{2})+(Ay^{2}) } =\sqrt{(29)^{2}+(14)^{2} }=32.20248438m[/tex]
Now in order to find the angle [tex]\alpha[/tex] we only need to use the trigonometry identity of tangent:
[tex]tan(\alpha )=\frac{Ay}{Ax}[/tex]
Isolating [tex]\alpha[/tex]
[tex]\alpha =arctan(\frac{Ay}{Ax} )=arctan(\frac{14}{29} )=25.76932762$^{\circ}$[/tex]
