Answer
$(a)\space 14.7\space cm$
$(b)\space 19.6^{\circ},\space south\space of\space west$
Work Step by Step
Here we use the component method to find the solutions. Let's take due east & due west as the positive directions.
(a) The resultant vector $R=(1)+(2)+(3)+(4)$
Due east component of = $R_{e}$
the resultant $R$
$R_{e}=(-27-23cos35^{\circ}-28cos55^{\circ}+35cos63^{\circ})\space cm$
$R_{e}=(-27-18.81+16.06+15.89)\space cm=-13.89\space cm$
Due north component of = $R_{n}$
the resultant $R$
$R_{n}=(0-23sin35^{\circ}-28sin55^{\circ}+35sin63^{\circ})\space cm$
$R_{n}=(-13.19-22.94+31.9)\space cm=-4.94\space cm$
By using the Pythagorean theorem,
The magnitude of $R=\sqrt {R_{e}^{2}+R_{n}^{2}}=\sqrt {(13.89\space cm)^{2}+(4.94\space cm)^{2}}=14.7\space cm$
(b) By using trigonometry (Please see the image),
$tan\theta=\frac{|R_{n}|}{|R_{e}|}=(\frac{4.94\space cm}{13.89\space})=19.6^{\circ},$ south of west
