Answer
$R=4N,
\alpha=60^{\circ}$ with negative x axis
Work Step by Step
Given that vector $a$ and $b$ and $\theta=60^{\circ}$
component along x axis
$f_{1}=2+4\cos{60}=2+2=4$ and $f_{2}={-6}$
net component in axis-x is $a=4-6=-2$
The component along the y axis is:
$f_{1}=4\sin60=2\sqrt{3}$ and $f_{2}=0$
The net component along the y-axis is $b=2\sqrt{3}$
The resultant magnitude of both vectors is:
$R=\sqrt{(a^2+b^2)}$
$R=\sqrt{((2\sqrt{3})^2+(-2)^2}=\sqrt{(12+{4})}=4N$
The angle $\alpha$ made in the west direction between the resultant and the x axis is:
$\alpha=\tan^{-1}{\frac{2\sqrt{3}}{{2}}}=60^{\circ}$
So:$
\alpha=60^{\circ}$ with negative x axis
