WTA Indian Wells: Angelique Kerber beats Jelena Ostapenko in first top ten win since return

Angelique Kerber has advanced to the third round of the WTA tournament in Indian Wells with a brilliant performance. The 36-year-old from Kiel defeated Jelena Ostapenko who is World Number 10 5-7, 6-3, 6-3 in round two.

This after 2:15 hours of play and trailing in sets. Kerber made her comeback on the WTA Tour in January after an 18-month maternity leave. She will next play Nao Hibino or Veronika Kudermetova in the third round.

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On Thursday, she won her first match aside from defeating Ajla Tomljanovic at the United Cup, she saw off Petra Martic from Croatia 6-3, 6-4 and was full of elation after sealing the win.

Despite her special start to the season, Ostapenko was prone to errors. And Kerber knew how to take advantage of that, although she conceded the first set despite having two set points of her own. She then went 2-0 up in the second set as she looked to cruise towards the win. But then Kerber turned on the heat, fought for every point and converted her fourth match point after 2:15 hours.

For Kerber, the so-called "Sunshine Double", the second part of which begins on March 20 in Miami, this win is super important and is a first top 10 victory in two years and four months. True passion, fight and belief personified. She is the lowest ranked player too to advance to the third round at Indian Wells. It is a first top 10 win since defeating Barbora Krejcikova at the 2021 Billie Jean King Cup. The former World No.1 back with a vengeance.

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her biggest win as a mom 👏 #TennisParadise

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