It looks like this spiral (or something very like it) can be drawn by segments connecting the following points: \begin{align} &(-r, r) \\\ &(r,-r) \\\ &(-r,-r) \\\ &(-2r,2r) \\\ &(2r,-2r) \\\ &(-2r,-2r) \\\ &(-3r,3r) \\\ &(3r,-3r) \\\ &(-3r,-3r) \\\ &(-4r,4r) \\\ &(4r,-4r) \\\ &(-4r,-4r) \\\ &\ldots \end{align}
That is, draw a segment from each point to the next one in the sequence. Choose a value of $r$ that makes the picture look right; if $r$ is too small you will just get a solid black triangle. The right value of $r$ may depend on the software you use.
Each "upward" line from $(-kr,-kr)$ to $(-(k+1)r, (k+1)r)$ makes an angle that is slightly to the left of "straight up." This is more evident in the inner part of the triangle, especially the upward segment from $(-r,-r)$ to $(-2r,2r).$ If you wanted that segment to be vertical (unlike the segments in the picture), you could replace each point that has coordinates $(-kr,-kr)$ with one at $(-(k+1)r,-kr).$