Answer
$\vec C= -( 4.9\;{\rm m})\;\hat i-( 1.89\;{\rm m})\;\hat j $
Work Step by Step
We told that
$$\vec A+\vec B+\vec C=0$$
and we need to find $\vec C$ in component form.
So,
$$\vec C=-\vec A-\vec B $$
$$\vec C=-(A_x\;\hat i+A_y\;\hat j)-(B_x\;\hat i+B_y\;\hat j) $$
$$\vec C=- A_x\;\hat i-A_y\;\hat j - B_x\;\hat i-B_y\;\hat j $$
$$\vec C=(- A_x- B_x)\;\hat i+(-A_y-B_y)\;\hat j $$
$$\vec C=-( A_x+ B_x)\;\hat i-( A_y+B_y)\;\hat j\tag 1 $$
We know that the horizontal component of any vector relative to the positive $x$-direction is given by
$$V_x=V\cos \theta$$
and the vertical component
$$V_y=V\sin\theta$$
where $\theta$ is the angle counterclockwise from the positive $x$-axis to the vector.
So,
$\theta_A=\bf 40^\circ $ while $\theta_B=360^\circ -20^\circ=\bf 340^\circ$
Plugging all that into (1);
$$\vec C=-( |\vec A|\cos \theta_A+ |\vec B|\cos \theta_B)\;\hat i-( |\vec A|\sin\theta_A+|\vec B|\sin\theta_B)\;\hat j $$
$$\vec C=-( 4\cos40^\circ + 2\cos 340^\circ)\;\hat i-( 4\sin40^\circ+2\sin340^\circ)\;\hat j $$
$$\boxed{\vec C= ( -4.9\;{\rm m})\;\hat i-( 1.89\;{\rm m})\;\hat j } $$
