Answer
a) See the figure below.
b)
$\vec A =(2.819\;{\rm m} )\hat i-(1.026\;{\rm m} )\hat j$,
$\vec B =(0\;{\rm m} )\hat i+(2\;{\rm m} )\hat j$,
$\vec C=-(1.71\;{\rm m} )\hat i-(4.698\;{\rm m} )\hat j$
c) $3.89\;\rm m, 286.6^\circ$
Work Step by Step
a) We drew the 3 vectors, as you see in the figure below.
b)
$$\vec A=A_x\hat i+A_y\hat j=(|A|\cos \theta_A )\hat i+(|A|\sin\theta_A )\hat j$$
Noting that $\theta_A=360-20=\bf 340^\circ$
Plugging the known;
$$\vec A =(3\cos 340^\circ )\hat i+(3\sin340^\circ )\hat j$$
$$\boxed{\vec A =(2.819\;{\rm m} )\hat i-(1.026\;{\rm m} )\hat j}$$
$$\vec B=B_x\hat i+B_y\hat j=(|B|\cos \theta_B )\hat i+(|B|\sin\theta_B )\hat j$$
Noting that $\theta_B =\bf 90^\circ$
Plugging the known;
$$\vec B =(2\cos 90^\circ )\hat i+(2\sin 90^\circ )\hat j$$
$$\boxed{\vec B =(0\;{\rm m} )\hat i+(2\;{\rm m} )\hat j}$$
$$\vec C=C_x\hat i+C_y\hat j=(|C|\cos \theta_C )\hat i+(|C|\sin\theta_C )\hat j$$
Noting that $\theta_C =180+70=\bf 250^\circ$
Plugging the known;
$$\vec C =(5\cos 250^\circ )\hat i+(5\sin 250^\circ )\hat j$$
$$\boxed{\vec C=-(1.71\;{\rm m} )\hat i-(4.698\;{\rm m} )\hat j}$$
c) The magnitude of $\vec D$ is found by applying the Pythagorean theorem.
$$|\vec D|=\sqrt{D_x^2+D_y^2}=\sqrt{\left(A_x+B_x+C_x\right)^2+\left(A_y+B_y+C_y\right)^2}$$
Plugging from above;
$$|\vec D| =\sqrt{\left(2.819+0 -1.71 \right)^2+\left( -1.026+2-4.698\right)^2}$$
$$|\vec D| =\color{red}{\bf 3.885}\;\rm m$$
and its direction is given by
$$\tan\alpha_D=\dfrac{D_y}{D_x}$$
Thus,
$$\alpha_D=\tan^{-1}\left[\dfrac{D_y}{D_x}\right]=\tan^{-1}\left[\dfrac{ -1.026+2-4.698}{2.819+0 -1.71}\right]=\bf -73.4^\circ$$
Thus,
$$\alpha_D=\color{red}{\bf 286.6^\circ}$$
