Answer
$26.6\ \mathrm{m}$.
Work Step by Step
Using:
Components of a vector, (3-5)
$ a_{x}=a\cos\theta$ and $ a_{y}=a\sin\theta$.
Magnitude and orientation, (3-6)
$a=\sqrt{a_{x}^{2}+a_{y}^{2}}$ and $\displaystyle \tan\theta=\frac{a_{y}}{a_{x}}$ ,
with
the angle between $\vec{A}$ and the $+x$ axis = $40^{\mathrm{o}}$
the angle between $\vec{C}$ and the $+x$ axis = $180^{0}+20.0^{0}=200^{\mathrm{o}}$.
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$B_{x}=C_{x}-\mathrm{A}_{x}=(15.0\mathrm{m})\cos 200^{0}-(12.0\mathrm{m})\cos 40^{\mathrm{o}}=-23.3\mathrm{m}$,
$B_{y}=C_{y}-\mathrm{A}_{y}=(15.0\mathrm{m})\sin 200^{0}-(12.0\mathrm{m})\sin 40^{\mathrm{o}}\\
=-12.8\mathrm{m}$.
$|\vec{B}|=\sqrt{(-23.3\mathrm{m})^{2}+(-12.8\mathrm{m})^{2}}=26.6\mathrm{m}$.
