American Kenin named WTA Player-of-the-Year
FLASHBACK: Sofia Kenin of the U.S. celebrates as she kisses the trophy after winning her match against Spain's Garbine Muguruza. (KIM HONG-JI/Reuters)
FLASHBACK: Sofia Kenin of the U.S. celebrates as she kisses the trophy after winning her match against Spain's Garbine Muguruza. (KIM HONG-JI/Reuters)

(REUTERS) – American Sofia Kenin has been named WTA Player-of-the-Year after winning her maiden Grand Slam singles title at the Australian Open. Kenin defeated World No.1 Ashleigh Barty in the semi-finals and twice Grand Slam champion Garbine Muguruza in the final to claim the title at Melbourne Park. Kenin also reached the French Open final and finished the season ranked a career-high No.4. The 22-year-old is the eighth American to win the WTA Player-of-the-Year accolade, joining Serena Williams, Martina Navratilova, Lindsay Davenport, Tracy Austin, Chris Evert, Venus Williams and Jennifer Capriati.

Iga Swiatek, who became the first Polish player to win a Grand Slam singles title when she beat Kenin in the French Open final, was named the WTA’s Most Improved Player-of-the-Year.
Former world number one Victoria Azarenka was named WTA Comeback Player-of-the-Year after she claimed her first title in over four years at the Western & Southern Open and advanced to her fifth career Grand Slam final at the U.S. Open. The Belarusian also regained her spot in the WTA Top 20. France’s Kristina Mladenovic and Hungary’s Timea Babos were voted WTA’s Doubles Duo-of-the-Year after capturing the Australian Open and Roland Garros titles.

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