Answer:
R=19.5m
[tex]\theta[/tex] = 4.65° S of W
Explanation:
Refer the attached fig.
displacement of the x and y components
x-component displacement is ([tex]R_{x}[/tex]) = [tex]A_{x}+B_{x}[/tex]
= A [tex]\sin[/tex](20°) + B [tex]\cos[/tex](40°)
= -12.0[tex]\sin[/tex](20°) + 20.0[tex]\cos[/tex](40°)
= -19.425m
x-component displacement is ([tex]R_{y}[/tex]) = [tex]A_{y}+ B_{y}[/tex]
= A [tex]\cos[/tex](20°) - B [tex]\sin[/tex](40°)
= 12.0[tex]\cos[/tex](20°) - 20.0[tex]\sin[/tex](40°)
= -1.579
resultant displacement
∴
R = [tex]\sqrt{R_{x}^{2} +R_{y}^{2} } }[/tex]
=[tex]\sqrt{(-19.425)^{2}+(-1.579)^{2} }[/tex]
=19.5m
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |[/tex]
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{1.579}{19.425} \right |[/tex]
[tex]\theta[/tex] = 4.65° S of W
