Tatiana Panova vs Anne Gaelle Sidot — Amelia Island Round of 64
Amelia Island · WTA · Round of 64
2 — 0
Sets
Monday, 5 April 1999 ·
clay
Panova T.7-6 7-6
Gaelle Sidot A.
Panova T.
772
Gaelle Sidot A.
660
Tatiana Panova defeated Anne Gaelle Sidot 7-6 7-6 in the Amelia Island round of 64 on clay. The result followed the form book — Tatiana Panova came in leading the head-to-head 2–0, defending last year's round of 64. Tatiana Panova extended the head-to-head to 3–0.
Key context
- Match story: Tatiana Panova won all two sets in tiebreaks against Anne Gaelle Sidot.
Head to head
2–0| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 2001-10-15 | Zurich | Q32 | hard | Tatiana Panova | 6-3 6-3 |
| 2001-10-08 | Filderstadt | Q16 | hard | Tatiana Panova | 6-1 7-6(18) |
Statistics
This season1st serve in·
61%
—
1st serve points won·
54%
—
2nd serve points won·
40%
—
Service points won·
49%
—
Return points won·
40%
—
Break points saved·
51%
—
Aces / match·
0.8
—
Double faults / match·
5.9
—
Averages over each player's completed matches in the selected window. ATP, WTA & Challenger.