Question 925315
The {{{n}}} angles of a polygon with {{{n}}} sides add up to {{{(n-2)*180^o}}} (see NOTE below).
So, for {{{n=7}}} , the sum of the angle measures is {{{5*180^o=900^o}}} ,
for {{{n=8}}} ,  the sum of the angle measures is {{{6*180^o=1080^o}}} , and
for {{{n=9}}} ,  the sum of the angle measures is {{{7*180^o=1260^o}}} .
The missing angle could measure
{{{1080^o-991^o=89^o}}} if it was a polygon with {{{8}}} sides,
and that is a very reasonable solution.
Wesley measured a polygon with 8 sides, an {{{highlight(octagon)}}} .
 
The numbers do not work out for less sides or more sides.
For {{{n<=7}}} , the sum of the angle measures would be at most {{{900^o}}} , which is less than {{{991^o}}} .
For {{{n>=9}}} , the missing angle would have to measure at least {{{1260^o-991^o=269^o}}} ,
and that would make it a convex polygon, something your teacher does not expect:
{{{drawing(300,300,-2,2,-2,2,
line(-1.5,-0.5,-1.5,0.5),line(1.5,-0.5,1.5,0.5),
line(-0.5,-1.5,0.5,-1.5),line(-0.5,1.5,0.5,1.5),
line(-0.5,-1.5,-1.5,-0.5),line(0.5,-1.5,1.5,-0.5),
line(-0.5,1.5,-1.5,0.5),line(0.5,1.5,0.518,0.5),
line(0.518,0.5,1.5,0.5),red(arc(0.518,0.5,1.2,1.2,0,269)),
locate(0,0.5,red(269^o))
)}}}

NOTE: A triangle has {{{3}}} sides, and {{{3}}} interior angles whose measures add up to {{{180^o}}} : {{{drawing(300,300,-1.5,1.5,-2,1,triangle(-1,0,1,0,0,-1.732))}}} .
Attaching another triangle with a common side you gain 1 extra side, and get a quadrilateral, like the kite below:
{{{drawing(300,300,-1.5,1.5,-2,1,triangle(-1,0,1,0,0,-1.732),
triangle(-1,0,1,0,0,0.8)
)}}} , with {{{3+1=4}}} sides, and {{{3+1=4}}} angles whose measures add up to {{{180^o+180^o=2*180^o}}} .
Repeating the process we get a pentagon,
with {{{3+2=5}}} sides, and {{{3+2=5}}} angles whose measures add up to {{{180^o+2*180^o=3*180^o}}} ,
and repeating the process once more we get a hexagon,
with {{{3+3=6}}} sides, and {{{3+3=6}}} angles whose measures add up to {{{180^o+3*180^o=4*180^o}}} :
{{{drawing(300,300,-1.5,1.5,-2,1,triangle(-1,0,1,0,0,-1.732),
triangle(-1,0,1,0,0,0.8),triangle(-1,0,0,-1.732,-1.1,-1)
)}}} and {{{drawing(300,300,-1.5,1.5,-2,1,triangle(-1,0,1,0,0,-1.732),
triangle(-1,0,1,0,0,0.8),triangle(-1,0,0,-1.732,-1.1,-1),
triangle(1,0,0,-1.732,1.1,-1)
)}}} Following the pattern, the {{{n}}} angles of a polygon with {{{n}}} sides add up to {{{(n-2)*180^o}}}
 
This is not a proof, but I thought it was a good way to remember the formula.