Answer
One puck makes an angle of $60^{\circ}$ north of east, and the other puck makes an angle of $60^{\circ}$ south of east.
Work Step by Step
By conservation of momentum, $p_x= 0$ and $p_y = 0$.
Let's assume that puck A moves west. Let $v$ be the speed of each puck.
$p_x = 0$
$0 = -m_Av+mv_{Bx}+mv_{Cx}$
$v = v_{Bx}+v_{Cx}$
$p_y = 0$
$0 = 0 +mv_{By}+mv_{Cy}$
$v_{By} = -v_{Cy}$
The vertical components of puck B and puck C have the same magnitude. Therefore, the horizontal components of puck B and puck C must have the same magnitude, otherwise they would have different speeds.
Then $v_{Bx} = \frac{v}{2}$ and $v_{Cx} = \frac{v}{2}$.
Since $\cos(60^{\circ}) = 0.5$, puck B and puck C must make an angle of $60^{\circ}$ with the +x-axis. Because their vertical components are in the opposite directions, one puck makes an angle of $60^{\circ}$ north of east, and the other puck makes an angle of $60^{\circ}$ south of east.
