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1969 U.S. Professional Indoor – Singles

Singles
1969 U.S. Professional Indoor
Champion Australia Rod Laver
Runner-up Australia Tony Roche
Final score 7–5, 6–4, 6–4
Events
Singles Doubles
← 1968 · U.S. Professional Indoor · 1970 →

Manuel Santana was the defending champion, but did not participate this year.

Rod Laver won the title, defeating Tony Roche 7–5, 6–4, 6–4 in the final.


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Wikipedia

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