解答:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/35a85edf8db1cb13743f5691de54564e93584bd9?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/71cf3bc79f3df8dcb16be513ce11728b461028d9?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
当点D′在矩形ABCD外部,点A′在△PDC内部,如图1,
∵∠BCD′=β,
∴∠DCD′=90°+β,
∵沿PB,PC将△PAB,△PDC翻折180°,得到△PA′B,△PD′C,
∴∠A′PB=∠APB=70°,∠DCP=∠DCP=
(90°+β),∠DPC=∠D′PC,
∴∠DPA′=180°-2×70°=40°,∠DPC=90°-∠DCP=
(90°-β),
∵∠A′PD′+∠DPA′=∠DPC+∠D′PC,
∴α+40°=2×
(90°-β),
∴β=50°-α;
同理得到当点D′在矩形ABCD外部,点A′在△PDC外部时,∠DPA′-∠A′PD′=∠DPC+∠D′PC,即40°-α=2×
(90°-β),
∴β=50°+α;
当点D′在矩形ABCD的内部,点A′在△PDC内部,如图2,
∵∠BCD′=β,
∴∠DCD′=90°-β,
∵沿PB,PC将△PAB,△PDC翻折180°,得到△PA′B,△PD′C,
∴∠A′PB=∠APB=70°,∠DCP=∠DCP=
(90°-β),∠DPC=∠D′PC,
∴∠DPA′=180°-2×70°=40°,∠DPC=90°-∠DCP=
(90°+β),
∵∠A′PD′+∠DPA′=∠DPC+∠D′PC,
∴α+40°=2×
(90°+β),
∴β=α-50°;
同理当点D′在矩形ABCD的内部,点A′在△PDC外部时,有β=α+50°,
综上所述,β=50°±α或β=α-50°.
故答案为β=50°±α或β=α-50°.