According to Ian Rapoport, the Jaguars are re-signing DE Dawuane Smoot on a two-year deal worth $14 million.
Smoot was one of Jacksonville’s best linemen last season, so it looks like the team decided to reward that.
Smoot, 26, was selected with the No. 68 overall pick by the Jaguars out of Illinois in 2017. He just finished the final year of a four-year rookie contract worth $3,364,160, which includes a signing bonus of $964,160.
In 2020, Smoot appeared in all 16 games for the Jaguars and recorded 25 total tackles, 5.5 sacks and two forced fumbles.
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