One person that comes to my mind is Kurt Angle but I don't really know if he's better than Chris Benoit.
2007-05-27 08:04:29 · 13 answers · asked by RKA: Leader of PAC 4 in Sports ➔ Wrestling
Goldberg is not near Benoit's level of wrestling!!! Even Chris Jericho can easily defeat Goldberg
2007-05-27 08:21:18 · update #1
hmmm...
MVP (thinks he is, anyways)
Dean Malenko
the Austin Starr
and, yes, Kurt Angle
2007-05-27 08:08:45 · answer #1 · answered by Leech 7 · 1⤊ 3⤋
Chris Benoit puts his body on the line every single match, I've seen that year after year... Perfectioning the cripple crossface, flying off the turnbukle breaking his nose I think three times... He's a great technical wrestler ,but he's also risky.. Thats what makes a great wrestler!! Taking chances for their fans!!
Not many out there better than Benoit... Maybe Jeff Hardy and (RIP) Eddie Guerrero... but really dunno who else.
2007-05-27 15:13:55 · answer #2 · answered by BTII 6 · 2⤊ 0⤋
The only wrestler greater than Chris Benoit, is Chris Benoit.
2007-05-27 15:07:20 · answer #3 · answered by tr0y 5 · 2⤊ 1⤋
Brett Hart
Goldberg
Kurt Angle
2007-05-27 15:15:14 · answer #4 · answered by Anonymous · 0⤊ 3⤋
Bret Hart
maybe the Undertaker
Sting in his prime
Kurt angle is a good techincal wrestler
and oh yeah john cena is better. well thats what vince thinks cause beniot had to tap out to queer cena
2007-05-27 15:09:33 · answer #5 · answered by Anonymous · 1⤊ 2⤋
Theres not alot.But Kurt Angle maybe....hmmm......Eddie Guerrero probably.....thats it really
2007-05-27 15:07:45 · answer #6 · answered by Anonymous · 1⤊ 2⤋
maybe.. Kurt angle.. but only because he has a gold medal....
2007-05-27 19:48:40 · answer #7 · answered by LION 6 · 0⤊ 0⤋
Hmmm.....theres not many...Maybe shawn micheals.
2007-05-27 15:08:55 · answer #8 · answered by Anonymous · 2⤊ 1⤋
I can do this ,,, I'm thinking now ,,,now I'm through
do you see my list
2007-05-27 15:12:47 · answer #9 · answered by jose 3 · 0⤊ 0⤋
either of the steiner brothers
2007-05-27 15:09:47 · answer #10 · answered by J S 4 · 0⤊ 2⤋