Question

如图一 在三角形abc中,o是角abc与角acb的平分线bo和co的交点

Answer 1

连接AO并延长AO交BC于D

∠BOD=∠ABO+∠BAD=∠BAD+1/2×∠B

∠COD=∠ACO+∠CAD=∠CAD+1/2×∠C

∴∠BOC=∠BOD+∠COD

=∠BAD+1/2×∠B+∠CAD+1/2×∠C

=∠A+1/2×(∠B+∠C)

=∠A+1/2×(180°-∠A)

=90°+1/2×∠A

∴当∠A=50°时,∠BOC=90°+25°=115°