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.
$\endgroup$ 11 Answer
$\begingroup$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$=7+14=21$
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