Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, [tex]A_x=A\text{Sin}37[/tex]
[tex]A_x=12[\text{Sin}(37)][/tex]
= 7.22 m
Vertical component, [tex]A_y=A[\text{Cos}(37)][/tex]
= 9.58 m
Similarly, Horizontal component of vector C,
[tex]C_x[/tex] = C[Cos(60)]
= 6[Cos(60)]
= [tex]\frac{6}{2}[/tex]
= 3 m
[tex]C_y=6[\text{Sin}(60)][/tex]
= 5.20 m
Resultant Horizontal component of the vectors A + C,
[tex]R_x=7.22-3=4.22[/tex] m
[tex]R_y=9.58-5.20[/tex] = 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,
[tex]R=\sqrt{(R_x)^{2}+(R_y)^2}[/tex]
= [tex]\sqrt{(4.22)^2+(4.38)^2}[/tex]
= [tex]\sqrt{17.81+19.18}[/tex]
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = [tex]\frac{\text{CB}}{\text{OB}}[/tex]
= [tex]\frac{R_y}{R_x}[/tex]
= [tex]\frac{4.38}{4.22}[/tex]
m∠COB = [tex]\text{tan}^{-1}(1.04)[/tex]
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
