解:设$AB = AC,$$BD\perp AC$于点$D,$分类讨论如下:
$①$当高与底边的夹角为$25°$时,高一定在$\triangle ABC$
的内部,如图①
∵$∠DBC = 25°,$$BD\perp AC,$∴$∠ADB = 90°$
又$∠ADB=∠DBC+∠C$
∴$∠C=∠ADB - ∠DBC = 65°$
又$AB = AC,$∴$∠ABC=∠C = 65°$
又$∠A+∠ABC+∠C = 180°$
∴$∠A = 180°-∠ABC-∠C = 50°$
$②$当高与另一腰的夹角为$25°,$且高在$\triangle ABC$
的内部时,如图②
∵$∠ABD = 25°,$$BD\perp AC,$∴$∠BDC = 90°$
又$∠BDC=∠A+∠ABD$
∴$∠A=∠BDC - ∠ABD = 65°$
∵$AB = AC,$∴$∠ABC=∠C$
又$∠A+∠ABC+∠C = 180°$
∴$∠C=∠ABC=\frac 12(180°-∠A)=57.5°$
$③$当高与另一腰的夹角为$25°,$且高在$\triangle ABC$
的外部时,如图③
∵$∠ABD = 25°,$$BD\perp AC,$∴$∠BDC = 90°$
又$∠BAC=∠BDC+∠ABD,$∴$∠BAC = 115°$
∵$AB = AC,$∴$∠ABC=∠C$
又$∠BAC+∠ABC+∠C = 180°$
∴$∠ABC=∠C=\frac 12(180°-∠BAC)=32.5°。$
综上,这个等腰三角形的各个内角的度数分别
为$65°,$$65°,$$50°$或$65°,$$57.5°,$$57.5°$
或$115°,$$32.5°,$$32.5°$