The resultant is 5 m at [tex]73.1^{\circ}[/tex]
Explanation:
To solve the problem, we have to resolve each of the two vectors into its component along the x-axis and the y-axis.
Vector A has magnitude 3 m and angle 20 degrees with the x-axis, so its components are
[tex]A_x = A cos \theta = (3)(cos 20)=2.82 m[/tex]
[tex]A_y = A sin \theta = (3)(sin 20)=1.03 m[/tex]
Vector B has magnitude 4 m and angle 110 degrees, so its components are
[tex]B_x = B cos \theta = (4)(cos 110)=-1.37 m[/tex]
[tex]B_y = B sin \theta = (4)(sin 110)=3.76 m[/tex]
So, the x- and y- components of the resultant vector are
[tex]R_x = A_x + B_x = 2.82+(-1.37)=1.45 m\\R_y = A_y+B_y = 1.03+3.76 =4.79 m[/tex]
Therefore the magnitude of the resultant is
[tex]R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.45)^2+(4.79)^2}=5.0 m[/tex]
While the direction is given by
[tex]\theta=tan^{-1}(\frac{R_y}{R_x})=tan^{-1}(\frac{4.79}{1.45})=73.1^{\circ}[/tex]
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