答案
解:(1)∵OD平分∠AOC,OE平分∠BOC,
∴∠DOC=
∠AOC,∠COE=
∠BOC
∴∠DOE=∠DOC+∠COE=
(∠BOC+∠COA)=
×(62°+180°-62°)=90°;
(2)∠DOE═
(∠BOC+∠COA)=
×(a°+180°-a°)=90°;
(3)∠DOA与∠COE互余;∠DOA与∠BOE互余;∠DOC与∠COE互余;∠DOC与∠BOE互余.
解:(1)∵OD平分∠AOC,OE平分∠BOC,
∴∠DOC=
∠AOC,∠COE=
∠BOC
∴∠DOE=∠DOC+∠COE=
(∠BOC+∠COA)=
×(62°+180°-62°)=90°;
(2)∠DOE═
(∠BOC+∠COA)=
×(a°+180°-a°)=90°;
(3)∠DOA与∠COE互余;∠DOA与∠BOE互余;∠DOC与∠COE互余;∠DOC与∠BOE互余.