Answer
a) $\vec B=-4\;\hat i+3\;\hat j$
b) $5,\;143.1^\circ$
Work Step by Step
a) From the given figure, it is obvious that
$$ \vec A + \vec B + \vec C =A_x\;\hat i+A_y\;\hat j+B_x\;\hat i+B_y\;\hat j+C_x\;\hat i+C_y\;\hat j=1\;\hat j$$
$$4\;\hat i+0\;\hat j+B_x\;\hat i+B_y\;\hat j+0\;\hat i-2\;\hat j=1\;\hat j$$
Thus,
$$\vec B=B_x\;\hat i+B_y\;\hat j=-4\;\hat i+3\;\hat j$$
$$\boxed{\vec B=-4\;\hat i+3\;\hat j}$$
b)
The magnitude of $\vec B$ is given by applying the Pythagorean theorem.
$$|\vec B|=\sqrt{B_x^2+B_y^2}=\sqrt{(-4)^2+3^2}=\color{red}{\bf 5}$$
and its direction is given by
$$\tan\alpha_B=\dfrac{B_y}{B_x}$$
Thus,
$$\alpha_B=\tan^{-1}\left[\dfrac{B_y}{B_x}\right]=\tan^{-1}\left[\dfrac{3}{-4}\right]=\bf -36.9^\circ$$
And since $\vec B$ is in the second quadrant,
$$\alpha_B=\color{red}{\bf 143.1^\circ}$$
