I am only able to make one possible case,
Where we take any 3-consecutive vertices, since one of the vertices contains angle of Heptagon, which is approximately 128.57°, we get 7 such triangles.
I am unable to any others cases where obtuse angle triangles are possible.
Consider all the triangle with the center of the circle circumscribing heptagon lying inside it.
Take any vertex of the $△ABC$ (let's say we take vertex $A$)
If $∠A=α$ then $∠BOC=2α$
But $∠BOC < 180°$
⇒ $∠A<90°$ (as $A$ and $O$ lie on the same side of $BC$)
Similarly we can prove $∠B<90°$ & $∠C<90°$
Using the same logic we can prove that the triangle is obtuse if $O$ lies outside the triangle.
Therefore for $△ABC$ to be obtuse the center O should lie outside the triangle.
There are only two non-congruent triangle satisfying the above property.
We can divide that into 2 cases (as shown in figure).
$\underline{CASE-1} $
Let $C$ be the vertex containing the obtuse angle.
$C$ can take any of the seven vertices.
So there are total of $7$ possibilities in this case.
$\underline{CASE-2} $
Let $C$ be the vertex containing the obtuse angle.
There are two triangles with $C$ as vertex & $C$ can take any of the seven vertices.
So there are total of $7×2=14$ possibilities in this case.
Therefore, the total number obtuse triangles that can be drawn$=7+14=21$