geometrypolygons
What is the smallest and largest possible angle of a triangle? (my guess = 1, 178)
What is the smallest and largest possible angle of a quadrilateral (convex or concave doesn't matter, and also perfect integers (no decimals) only)? (my guess = 1, 357)
If my guesses are right, then:-
The smallest possible angle of an n-sided polygon is always 1.
The largest possible angle of an n-sided polygon is always ((n – 2) * 180) – (n – 1)
Angles must be specified in degrees not radians.
The polygon with the least possible area :-
Conditions
The polygon must contain the 1*1 square wholly
The polygon must be contained in the 2 by 2 square in a way that this square is the smallest square containing the polygon
Deduction
These conditions lead us to something like this:
All these convex polygons have the same area = 1.5 units , because triangles with same base and lying between same parallel lines have equal area, as their height is also equal
If we consider the polygon given in possibility 3 or 4 below, these polygons already have 1.5 sq. units area.
In possibility 3 (below), $\frac{BE*EA}{2}= 0.75$ Therefore total area = 1.75
In possibility 4 (below), $\frac{CE*EA}{2}= 1$ Therefore total area = 2
Therefore, the polygons with least area = the ones in the first picture, with area = 1.5 sq. units
NOTE
Regarding the NOTE in your question, Even if the polygon is concave, its minimum area needs to be more than 1 sq. units.
Three angles will always be 90 or more than 90, so we focus on the 4th angle.
Possibility 1,
The angle has to be 45 (consider the triangle ABC in first figure. if one angle is 90, other 2 must be 45)
Possibility 2
$AD=\frac{1}{4}units$
In isosceles triangle ABC, BC = 1 and AD=$\frac{1}{4}$
Let $\angle BAE= \angle A$
$\tan A = \frac{BE}{AE} = 0.5$
$A = \tan^{-1}0.5$ ; $A=26.5650 $
$2A = \angle CAB = 53.130$
We see that when the triangle in possibility 2 that when the triangle in the polygon lying outside the small square is DBC, the angle is 45. As the triangle transforms, or as we move the apex of the triangle to a point halfway between D and F, the angle starts increasing until finally it is 53.13 at A. then again it starts decreasing till it reaches F, it again becomes 45. What happens next?
Possibility 3
The angle formed on the side of the square starts decreasing to below 45
here, $AD=\frac{3}{4}units$
$\tan \angle CAG = \frac{\frac{1}{4}}{2} = \frac{1}{8} =0.125$
$\angle CAG= \tan^{-1} 0.125= 7.1256$
and
$\tan \angle BAE=\frac{1.5}{1}=1.5$
$\angle BAE= \tan^{-1}1.5 = 56.3099$
Therefore
$ \angle BAC = 56.3099-7.1256 = 49.1843 $
Possibility 4
Again here, the angle further decreases, till it reaches A. Now, the angle's value is least of all.
In right triangle ADC, CD = $\frac{\sqrt 2}{2}$ and AD= $\sqrt 8 - \frac{\sqrt 2}{2}$
$\tan A = \frac{\frac{\sqrt 2}{2}}{\sqrt 8 - \frac{\sqrt 2}{2}}$ ($\angle A=\angle CAD$)
$\tan A= 0.3333...$
$A = \tan^{-1}0.333...$ ; $A=18.4349 $
$2A = \angle CAB = 36.86989764$
For an n-gon, considering a line rotating around it rotates 360. This implies the standard result that the sum of the exterior angles in the direction of the rotation is 360 so the sum of the interior angles is n*180-360 = 180(n-2).
Therefore the sum of the exterior angles on both sides of each vertex is 720.
From your first diagram, each angle of the extended polygon is 180 minus the sum of the adjacent exterior angles, so their sum is 180*n-720 =180(n-4).
Best Answer
The set of values which are valid angle measurements does not have a minimum or a maximum if we are talking about non-degraded convex polygons. It does have an infimum and supremum - in case of a triangle the infimum is 0 and the supremum is 180.