Los Angeles Angels’ Mike Trout named MLB’s American League MVP

CHICAGO — While the Los Angeles Angels stumbled, Mike Trout soared again.

Just too good to ignore.

Trout won the American League Most Valuable Player award Thursday for the second time in three years in balloting by the Baseball Writers’ Association of America.

“To win it one time, it’s hard to do,” Trout said. “Twice, I guess you saw my emotions tonight, it was something special.”

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).

“At the end of the season a lot of people were asking me about the same question, you know, if your team has a losing record is it going to hurt you in the MVP?” Trout said. “With this MVP this year, I guess it doesn’t matter.”

Voting by BBWAA members was completed by the start of the playoffs.

Trout 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 finished sixth in his final year in the majors.

“A well-deserved honor for a player that puts his name in the MVP conversation year in and year out,” Angels manager Mike Scioscia said in a release. “Mike is the consummate team player and deserves every accolade he receives on the field just as he should off of it for the person he is.”