  We show that any planar straight line graph (PSLG) with $n$ vertices has a
conforming triangulation by $O(n^{2.5})$ nonobtuse triangles (all angles $\leq
90^\circ$), answering the question of whether any polynomial bound exists. A
nonobtuse triangulation is Delaunay, so this result also improves a previous
$O(n^3)$ bound of Eldesbrunner and Tan for conforming Delaunay triangulations
of PSLGs. In the special case that the PSLG is the triangulation of a simple
polygon, we will show that only $O(n^2)$ triangles are needed, improving an
$O(n^4)$ bound of Bern and Eppstein. We also show that for any $\epsilon >0$,
every PSLG has a conforming triangulation with $O(n^2/\epsilon^2)$ elements and
with all angles bounded above by $90^\circ + \epsilon$. This improves a result
of S. Mitchell when $\epsilon = 3 \pi /8 = 67.5^\circ $ and Tan when $\epsilon
= 7\pi/30 =42^\circ$.
