Carioca advanced to the stage after beating Tommy Paul and will face the Italian, owner of four Grand Slam titles
The Brazilian João Fonseca will face the number two in the worldJannik Sinner, in the round of 16 at Indian Wells. The duel was defined after the Rio’s victory over Tommy Paul by 2 sets to 0with splits of 6/2 and 6/3, in the early hours of this Monday (9), in California.
Sinner secured his place in the next round by defeating Canadian Denis Shapovalov in a match played earlier on Sunday (8). Current number two in the world rankings and owner of four Grand Slam titles, the Italian is still looking to win the tournament trophy in Indian Wells for the first time.
In the match that guaranteed classification, João Fonseca started better and confirmed the first serve before breaking Paul’s serve, opening 2/0. The Brazilian maintained regularity in the ball exchanges and extended it to 3/0. Then, he won a game again on his opponent’s serve and reached 4/1 on the scoreboard.
Paul managed to confirm a game and reduce the difference, but Fonseca maintained control of the set. With a new break, he opened 5/2 and closed the partial at 6/2. After the end of the first set, the Brazilian asked for medical attention before the start of the sequence.
In the second half, Paul started better and won the first game on Fonseca’s serve, opening 2/0. The Brazilian reacted, got the break and tied the set at 2/2. From then on, he started to confirm his service games and maintained the balance of the dispute.
With the score at 4/3, Fonseca broke the American’s serve again after a long exchange of balls and opened 5/3. He then closed the set at 6/3, guaranteeing qualification to the round of 16.
*With information from Estadão Conteúdo
