Answer
The momentum of the daughter nucleus is directed at an angle of $~~28.1^{\circ}~~$ above the positive x axis.
Work Step by Step
Note that the initial momentum is 0.
We can use conservation of momentum to find the momentum $p_n$ of the daughter nucleus:
$p_f = p_i$
$p_n+(-1.2\times 10^{-22}~kg~m/s)~\hat{i}+(-6.4\times 10^{-23}~kg~m/s)~\hat{j} = 0$
$p_n = (1.2\times 10^{-22}~kg~m/s)~\hat{i}+(6.4\times 10^{-23}~kg~m/s)~\hat{j}$
We can find the direction of the momentum of the daughter nucleus as an angle $\theta$ above the positive x axis:
$tan~\theta = \frac{6.4\times 10^{-23}~kg~m/s}{1.2\times 10^{-22}~kg~m/s}$
$\theta = tan^{-1}~(\frac{6.4\times 10^{-23}~kg~m/s}{1.2\times 10^{-22}~kg~m/s})$
$\theta = tan^{-1}~(0.533)$
$\theta = 28.1^{\circ}$
The momentum of the daughter nucleus is directed at an angle of $~~28.1^{\circ}~~$ above the positive x axis.
