Question

如图,D为△ABC内一点,∠ABD=20°,∠ACD=25°,∠A=35°,求∠BDC的度数

Answer 1

∵∠A+∠ABC+∠ACB=180°

∴∠ABC+∠ACB=180°-35°=145°

∵∠ABC=∠ABD+∠DBC

∠ACB=∠ACD+∠DCB

∴∠ABD+∠DBC+∠ACD+∠DBC=∠ABC+∠ACB=145°

即∠DBC+∠DCB=145°-∠ABD-∠ACD=145°-20°-25°=100°

∴∠BCD=180°-(∠DBC+∠DCB)=180°-100°=80°