Three vectors are shown in the figure. Their magnitudes aregiven in arbitrary units. Determine the sum of the three vectors.Give the resultant in terms of•components•magnitude•angle with the +x axis(Figure 1)

Components:
First, calculate the cartesian components of each vector, as follow:
[tex]\begin{gathered} A_x=44.0\cdot\cos (28.0)=38.85 \\ A_y=44.0\cdot\sin (28.0)=20.65 \\ B_x=-26.5\cdot\cos (56.0)=-14.82 \\ B_y=26.5\cdot\sin (56.0)=21.97 \\ C_x=0 \\ C_y=-31.0 \end{gathered}[/tex]Next, consider that the components of the resultant vector R, are given by the sum of the x components and y components of all vectors A, B and C:
[tex]\begin{gathered} R_x=A_x+B_x+C_x=38.5-14.82+0=23.68 \\ R_y=A_y+B_y+C_y=20.65+21.97-31.0=11.62 \end{gathered}[/tex]Magnitude:
The magnitude is calculated as follow:
[tex]R=\sqrt[]{R^2_x+R^2_y}=\sqrt[]{(23.68)^2+(11.62)^2}=26.38[/tex]Angle with x axis:
The angle related to the x axis is obtained as follow:
The tangent of the angle related to the x axis is:
[tex]\tan \theta=\frac{R_y}{R_x}[/tex]which is basically, the quotient between the opposite site and adjacent side of a right triangle formed by the components of the vector.
To obtain the angle you apply tan^-1 to cancel out the tangent, as follow:
[tex]\theta=\tan ^{-1}(\frac{R_y}{R_x})=\tan ^{-1}(\frac{11.62}{23.68})=26.14\degree[/tex]