Answer
a) $36.1^{\circ}$ north of west
b) $36.1^{\circ}$ south of west
Work Step by Step
a) We find:
$F^{2}=A^{2}+B^{2}$
$F=\sqrt {A^{2}+B^{2}}$
$F=\sqrt {(445N)^{2}+(325N)^{2}}=551N$
$tan\theta=\frac{B}{A}$
$\theta=tan^{-1}\frac{325N}{445N}=36.1^{\circ}$ north of west
b) We find:
$F^{2}=A^{2}+(-B)^{2}$
$F=\sqrt {A^{2}+(-B)^{2}}$
$F=\sqrt {(445N)^{2}+(-325N)^{2}}=551N$
$tan\theta=\frac{B}{A}$
$\theta=tan^{-1}\frac{325N}{445N}=36.1^{\circ}$ south of west
