Question 703883
For a polygon where all the angles have the same measure
(like an equilateral triangle, a square, a rectangle, a regular hexagon, etc)
the measure of an exterior angle equals
{{{360^o}}} divided by the number of sides.
So for an equilateral triangle,
each exterior angle measures {{{360^o/3=highlight(120^o)}}}
{{{drawing(300,200,-0.6,1.2,-0.15,1.05,
triangle(-0.5,0,0.5,0,0,0.866),
red(arrow(0.5,0,1,0)),red(arrow(0.5,0,0.25,0.433)),
locate(0.52,0.15,red(120^o))
)}}}
 
Here's why:
When you are drawing a polygon,
at each vertex, to "turn the corner",
you have to change direction by a certain angle.
That angle if the exterior angle.
For a square, you turn {{{90^o}}} at each corner,
and if you keep going,
as you are retracing the first side,
you have turned {{{90^o}}} four times,
for a total of {{{4(90^o)=360^o}}},
and you are going in the same direction as when you started.
For any regular polygon of {{{n}}} sides,
the {{{n}}} exterior angles have the same measure {{{A}}},
and {{{n(A)=360^o}}}, so {{{A=360^o/n}}}