The Tampa Bay Buccaneers locked up one of their most important defenders for the future just a day before their regular-season finale against the Carolina Panthers. On Saturday evening, the Bucs announced that they had agreed to a four-year extension with defensive tackle Tevita Tuliʻakiʻono Tuipulotu Mosese Vaʻhae Fehoko Faletau Vea — better known as Vita Vea. According to NFL Media’s Ian Rapoport, the four-year extension is worth over $73 million.
Vea is in the midst of his fourth season with the Buccaneers, and has recorded 31 combined tackles, 12 quarterback hits and a career-high 4.0 sacks in 15 games played. He was originally selected by Tampa Bay with the No. 12 overall pick in the 2018 NFL Draft out of Washington, where he was named the Pat Tillman Pac 12 Defensive Player of the Year back in 2017. In 49 career games played, the 6-foot-4, 347-pound defenseman has recorded 104 combined tackles, 11.5 sacks, 31 QB hits, 16 tackles for loss and six passes defensed.
This reported deal puts Vea into the top five of highest-paid defensive tackles, according to Spotrac. While the deal is reportedly worth over $73 million, that number over four years has an AAV of $18.25 million, which ranks fourth in the NFL ahead of Jonathan Allen of the Washington Football Team, and behind Chris Jones of the Kansas City Chiefs.