The Los Angeles Rams are in desperate need of outside pass-rush help and late Sunday night, they took the first step toward addressing it. They agreed to terms on a one-year deal with Dante Fowler Jr., keeping him in Los Angeles for another season and preventing him from hitting free agency on Wednesday.
Giving him a one-year contract to allow him to prove himself is a smart move, considering his lack of overall production in the NFL and his limited time in the Rams’ system. He played just 11 total games with L.A. last season, recording 3.5 sacks and eight tackles for loss.
That one-year deal won’t come cheap, though. According to Ian Rapoport, Fowler’s contract is worth $14 million for just the 2019 season.
As Rapoport states, that’s about the value of what would’ve been the fifth-year option for Fowler. The Rams didn’t have the chance to exercise it because they acquired him after the Jaguars had already declined it, but they clearly feel good about paying him a significant amount of money.
The exact structure of the deal is unknown up to this point, but considering it’s only for one season, it’s highly likely that most of the money is guaranteed. The Rams may have built incentives into the contract to make it worth up to $14 million, but either way, Fowler is going to get paid in 2019.
Los Angeles could’ve placed the franchise tag on Fowler and paid him $15.4 million, which wouldn’t have been much more to guarantee him staying for another year. However, the Rams were able to save some money by locking him up to a one-year deal, albeit an expensive one.
