N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\angle EFD=90^\circ$, then how many degrees are in the measure of $\angle FDE$? [asy]
size(5.5cm);
pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); }

pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216);
pair f=e-(0,2*sin(pi/5)*sin(pi/10));
dot(a); dot(b); dot(c); dot(d); dot(e); dot(f);
label("$A$",a, WNW);
label("$B$",b, ENE);
label("$C$",c, E);
label("$D$",d, ESE);
label("$E$",e, W);
label("$F$",f, WSW);
draw(d--f--a--b--c--d--e);
draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0));
[/asy]

18

Step-by-step explanation:

Each interior angle of a regular pentagon is 108 degrees. So Angle AED is 108 degrees. Since Angle AEF is a straight line (180 degrees), Angle FED is 72. This is because 180-108 = 72. Now, since a triangle has a total of 180 degrees, we add 72 and 90, because those are the 2 degrees we have calculated. This gives us a total of 162. Now, we subtract 162 from 180 to find out the degree of Angle FDE. This is 18. So our final answer is 18.

Sidenote: I hope this answer helps!

hi there! my name is zalgo and i am here to you out today. the answer is pretty simple (in a way). this equation equals to 25.2!

i hope this answer !

"stay and stay proud! " - zalgo