解:设∠A=x,
∵AP1=P1P2=P2P3=…=P13P14=P14A,
∴∠A=∠AP2P1=∠AP13P14=x,
∴∠P2P1P3=∠P13P14P12=2x,
∴∠P3P2P4=∠P12P13P11=3x,
…,
∠P7P6P8=∠P8P9P7=7x,
∴∠AP7P8=7x,∠AP8P7=7x,
在△AP7P8中,∠A+∠AP7P8+∠AP8P7=180°,
即x+7x+7x=180°,
解得x=12°,
即∠A=12°.
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