1.设∠B=b,∠ADE=a,△ADE中,2a+(150-2b)=30,a-b=15;△ADC中,a+X+b+(150-2b)=180,得a-b+x=30,a-b=15于是x=15
2.∠A=a,∠BOC=2a-5,延长BO交ac于K,角KOC=175-2a,∠omc=2a-30,∠bma=210-2a,(210-2a)+a+25=180,a=55
3.设∠C=b,∠DAC=a,根据各种等腰,得2a+b=180°,3b+a=180du,得a=72°,b=36°,于是∠B=∠C=36°∠BAC=108°