Answer
a) $=1.14\times10^{-22}kg\frac{m}{s}$
b) $147^o$ from the electron and $123^o$ degrees from the neutrino.
Work Step by Step
Assume the radioactive nucleus is located at the origin, the electron went in the North direction and the neutrino went in the East direction.
a) $\sqrt{(9.6\times10^{-23}kg\frac{m}{s})^2+(6.2\times10^{-23}kg\frac{m}{s})^2}$
$=1.14\times10^{-22}kg\frac{m}{s}$
b) $\arctan\Big(\frac{9.6\times10^{-23}kg\frac{m}{s}}{6.2\times10^{-23}kg\frac{m}{s}}\Big)=57.1^o$
The direction is $180^o+57.1^o=237^o$
The direction of the electron is $90^o$ so the direction of the recoiling nucleus was $237^o-90^o=147^o$ from the electron and $360^o-147^o-90^o=123^o$ degrees from the neutrino.