Last Updated on January 19, 2019
The ninth racket of the world, Japanese tennis player Kei Nishikori, in the third round of the Australian Open outplayed Portuguese Joao Sousa, who occupies 44th place in the world ranking, – 7/6 (2), 6/1, 6/2.
For 2 hours and 6 minutes of the match, Kei made three aces and five breaks, having erred three times when serving. Whereas João filed through nine times and implemented one break point, making one double mistake.
Through personal meetings of opponents was 3: 1 in favor of the Japanese. In the fourth round of the Australian Open will play with the Spaniard Pablo Carreño-Busta.
