Question

如右图,△ABC中,∠A=62°,∠ABC和∠ACD的角平分线BO和CO相交于点O,则∠BOC的度数等 求详解

Answer 1

∵，∠ABC的平分线和∠ACD的外角平分线相交于点O
∴∠OBC=1/2 ∠ABC
∠OCD=1/2∠ACD
又∠A +∠ABC =∠ACD
∴ 1/2（∠A +∠ABC） =1/2∠ACD
即1/2∠A +1/2∠ABC =1/2∠ACD
∴31°+ ∠OBC=∠OCD
∴ ∠BOC=∠OCD-∠OBC=31°

Answer 2

31°
∠ACD=∠A+∠ABC

∠OCD=0.5∠ACD=∠OBC+∠O=0.5∠ABC+∠O

所以∠O=0.5∠A

Answer 3

boc=180-obc-bco
=180-0.5abc-bca-31-0.5abc
=180-abc-bca-31
=31

Answer 4

∠DCO=∠BOC ∠OBC
∠ACO ∠DCO=∠A ∠ABO ∠OBC
2∠DCO=62° 2∠OBC
2(∠BOC ∠OBC)=62° 2∠OBC
2∠BOC=62° ∠BOC=31°