Question 917766
The external angles are the angles you have to turn at each corner as you go around the polygon.
 
Naturally, after you complete one turn around the polygon, you are facing the same way as when you started, so you have turned a total angle of {{{360^o}}} .
The sum of the measures of all those external angles is {{{360^o}}} .
Your teacher (or the book) had to tell you that, but isn't it obvious now that I put it that way?
 
Let's say {{{n}}}= number of angles measuring {{{27^o}}} . Then,
{{{55^o+35^0+n*27^o=360^o}}}--->{{{90^0+n*27^o=360^o}}}--->{{{n*27^o=360^o-90^o}}}--->{{{n*27^o=270^o}}}--->{{{n=270^o/27^o}}}--->{{{n=10}}} .
That tells you that the polygon had the two external angles measuring {{{55^o}}}  and {{{35^o}}} ,
plus {{{10}}} angles measuring {{{27^o}}} each,
for a total of {{{2+10=12}}} external angles.
A polygon with {{{12}}} external angles must have {{{12}}} internal angles and {{{highlight(12)}}} sides.
 
NOTE:
I like to assume that the polygon is a convex polygon, so you are always turning in the same direction (always veering right, or always turning to your left).
Otherwise, you would have to measure turns in one direction as positive angles, and turns in the other direction as negative angles. Then, adding angles gets complicated.