What is the maximum possible number of acute interior (vertex) angles in a pentagon?
The sum of the interior angles of any polygon is
(N-2)180°
where N is the number of sides (and interior angles).
In the case of a pentagon, N=5, so the sum of the
5 interior angles is
(5-2)180° = (3)180° = 540°.
Let the angles be A, B, C, D, and E. Then
A + B + C + D + E = 540
Let's see if all 5 can be acute:
A < 90°
B < 90°
C < 90°
D < 90°
E < 90°
Adding all those inequalities:
A < 90°
B < 90°
C < 90°
D < 90°
E < 90°
---------------
A+B+C+D+E < 450°
No, since that sum must equal 540°,
not less that 450°. But we could make
4 of them, A, B, C, and D acute,
A < 90°
B < 90°
C < 90°
D < 90°
---------------
A+B+C+D < 360°
and so E would then have to be more than 180°,
which is called a "reflex angle".
for instance
Edwin
