Raonic makes Brisbane tennis final

Canada’s Milos Raonic will play in Sunday’s Brisbane International final after emerging victorious in an epic semi-final against Kei Nishikori.

The world No.8 held his nerve in a 151-minute match which didn’t feature a single break of serve, defeating Nishikori 6-7 (4-7) 7-6 (7-4) 7-6 (7-4).

Raonic entered the match leading the tournament with 32 aces from his two prior matches and added a whopping 34 to his tally in this match.

It’s the second time Raonic has beaten Nishikori in six meetings between the pair, with the Japanese world No.5 defeated in Brisbane at the semi-final stage for the third straight year.

Nishikori, last year’s US Open finalist, also bows out having not dropped a service game in the tournament.

Raonic will now play the winner of Saturday’s second semi-final between fourth seed and world No.11 Grigor Dimitrov and 17-time grand slam champion Roger Federer.

In the longest match yet in the tournament, Nishikori twice had break point opportunities in the first set but could not convert as Raonic used his booming serve to bail him out of trouble.

There was no way out for the Canadian in the tiebreak however, Nishikori making no mistake when he got the mini-break to wrap up the set in 49 minutes.

The pair again couldn’t be separated in the second set, with Raonic failing to convert three break points.

Another tiebreak ensued but Nishikori was the one who blinked this time, double-faulting on the ninth point of the tiebreak.

Raonic took full advantage, blasting the next two points off his serve to level the match.

Nishikori failed to convert another break point in the third set and made a crucial error midway through the final set tiebreak to hand Raonic an advantage he wouldn’t relinquish.

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