23.38 ° to the positive x-axis
Explanation
Step 1
draw the vectors
Step 2
add the vectors, ( component to component)
so
a) get the components of V1
[tex]\begin{gathered} 23.87\text{ at 0\degree} \\ so \\ V_{1x}=23.87\text{ cos 0=23.87} \\ V_{1y}=23.87\text{ sin 0= 0} \end{gathered}[/tex]b) components of V2
[tex]\begin{gathered} 12.9\text{ at 90 \degree} \\ so \\ V_{2x}=\text{ 12.9 cos 90=0} \\ V_{2y}=12.9\text{ sen 90=12.9} \end{gathered}[/tex]c) add
[tex]\begin{gathered} Vdis_x=23.87+0=23.87 \\ Vdis_y=0+12.9 \end{gathered}[/tex]so, the magnitude of the displacement is
[tex]\begin{gathered} \lvert Vd\rvert=\sqrt[]{V^2_x+V^2_y} \\ replace \\ \lvert Vd\rvert=\sqrt[]{23.87^2+12.9^2} \\ \lvert Vd\rvert=27.132 \end{gathered}[/tex]finally, the direction
as we have a rigth triangle
let
opposite side= 12.9
adjacent side=23.8
Now, use the tan function to find the angles
so
[tex]\begin{gathered} \tan \emptyset=\frac{opposite\text{ side }}{\text{adjancent side}} \\ \text{replace} \\ \tan x=\frac{12.9}{23.87} \\ \text{isolate x} \\ x=\tan ^{-1}(\frac{12.9}{23.87}) \\ x=28.38 \end{gathered}[/tex]therefore, the direction of the resultant is
23.38 ° to the positive x-axis
I hope this helps you
