Question 234283
What is the maximum possible number of acute interior (vertex) angles in a pentagon?

<pre><font size = 4 color = "indigo"><b>
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

{{{drawing(400,260,-1.3,1.3,-.3,1.7,

line(-1,0,1,0), line(1,0,.7,1.4), line(.7,1.4,0,1),

line(0,1,-.7,1.4), line(-.7,1.4,-1,0),

locate(-.7,1.5,A), locate(-1,0,B), locate(1,0,C),
locate(.7,1.5,D), locate(0,1,E) 

 )}}} 
Edwin</pre>