Robert Howe (tennis)
Australian tennis player / From Wikipedia, the free encyclopedia
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Robert (Bob) Howe (3 August 1925 – 30 November 2004)[2] was an Australian tennis player. Although he won 29 singles titles, his main successes were achieved in the doubles competition. He won four mixed doubles Grand Slam titles, including the Wimbledon mixed doubles championship in 1958.[3]
Quick Facts Full name, Country (sports) ...
Full name | Robert Neville Howe |
---|---|
Country (sports) | Australia |
Born | (1925-08-03)3 August 1925 Sydney, NSW, Australia |
Died | 30 November 2004(2004-11-30) (aged 79) Santa Ana, California, US |
Height | 1.77 m (5 ft 10 in) |
Turned pro | 1968 (amateur from 1949) |
Retired | 1971 |
Plays | Right-handed (two-handed backhand)[1] |
Singles | |
Career record | 500-369 |
Career titles | 29 |
Grand Slam singles results | |
Australian Open | QF (1958, 1963) |
French Open | 3R (1957) |
Wimbledon | 4R (1956, 1962, 1965) |
US Open | 4R (1957) |
Doubles | |
Career record | 208–94 |
Career titles | 18 |
Highest ranking | No. 12 (30 August 1977) |
Grand Slam doubles results | |
Australian Open | F (1959) |
French Open | F (1958, 1961) |
Wimbledon | SF (1956, 1957) |
Mixed doubles | |
Grand Slam mixed doubles results | |
Australian Open | W (1958) |
French Open | W (1960, 1962) |
Wimbledon | W (1958) |
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