A stationary block explodes into three equal parts. Two of the...

Let the total mass of the stationary block be $3m$. Since it explodes into three equal parts, the mass of each part is $m$.

The initial momentum of the system is zero because the block is stationary.

Let the velocities of the three parts be $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$.
Given that two parts travel along the $X$ and $Y$ axes with a speed of $10\text{ m/s}$, we have:
$\vec{v}_1 = 10\hat{i}\text{ m/s}$
$\vec{v}_2 = 10\hat{j}\text{ m/s}$

According to the law of conservation of linear momentum:
$\vec{P}_{\text{initial}} = \vec{P}_{\text{final}}$
$0 = m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3$
$0 = m(10\hat{i}) + m(10\hat{j}) + m\vec{v}_3$
$\vec{v}_3 = -10\hat{i} - 10\hat{j}$

The magnitude of the velocity of the third part is:
$|\vec{v}_3| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2}\text{ m/s}$

The direction of $\vec{v}_3$ is given by the angle it makes with the positive $X$-axis. Since both the $x$ and $y$ components are negative, the vector lies in the third quadrant.
The angle made with the negative $X$-axis is \tan^{-1}\left(\dfrac{10}{10}\right) = 45^{}^{{}^{\circ}}.
Thus, the angle from the positive $X$-axis is 180^{}^{{}^{\circ}} + 45^{}^{{}^{\circ}} = 225^{}^{{}^{\circ}} counterclockwise.
This is equivalent to 135^{}^{{}^{\circ}} clockwise, meaning it is directed 135^{}^{{}^{\circ}} below the positive $X$-axis.