Angels' Mike Trout Named 2016 American League MVP | NBC 10 Philadelphia

Angels' Mike Trout Named 2016 American League MVP

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    Mike Trout of the Los Angeles Angels of Anaheim celebrates after hitting a lead off home run in the first inning of the 86th MLB All-Star Game at the Great American Ball Park on July 14, 2015 in Cincinnati, Ohio.

    Los Angeles Angels outfielder Mike Trout won his second career American League Most Valuable Player award on Thursday.

    While the Angels finished fourth in the AL West, Trout was his usual brilliant self. The center fielder batted .315 with 29 homers, 100 RBIs and 30 steals. He scored 17 percent of Los Angeles' runs, the highest percentage for an AL player since Rickey Henderson with the 1985 New York Yankees.

    Trout, who was a unanimous winner in 2014, had finished second in three of the past four years. He becomes the first MVP from a losing team since Alex Rodriguez for Texas in 2003 and just the fifth player ever to accomplish the feat, joining Hall of Famers Ernie Banks (1958 and 1959), Andre Dawson (1987) and Cal Ripken (1991).

    "It's an unbelievable feeling," Trout said. "Just trying to get better every year."

    Trout, 25, received 19 first-place votes and 356 points. Mookie Betts, who batted .318 with 31 homers, 113 RBIs and 26 steals in 158 games for the Boston Red Sox, was second with 311 points, and AL batting champion Jose Altuve of Houston was third. Retiring Red Sox slugger David Ortiz got one first-place vote and finished sixth in his final year in the majors.