Answer
$13.3\space m,\space 51.7^{\circ}$
Work Step by Step
Please see the attached image first.
Here we use the component method to find the magnitude & the direction of the resultant vector R.
We can write,
$R=A+B+C+D$
x component of $R=R_{x}=Dcos50^{\circ}-Acos20^{\circ}-Ccos35^{\circ}$
$R_{x}=26\space m\times0.64-16\space m\times0.94-12\space m\times0.82=-8.24\space m$
y component of $R=R_{y}=B-Dsin50^{\circ}+Asin20^{\circ}-Csin35^{\circ}$
$R_{y}=11\space m-26\space m\times0.77+16\space m\times0.34-12\space m\times0.57=-10.42\space m$
$R=\sqrt {R_{x}^{2}+R_{y}^{2}}=\sqrt {(-8.24\space m)^{2}+(10.42\space m)^{2}}=13.3\space m$
From the right-hand side figure of the image,
$tan\theta=\frac{|R_{y}|}{|R_{x}|}=\frac{10.42}{8.24}=\gt \theta=tan^{-1}(1.26)=51.7^{\circ}$
