在锐角三角形abc中,bd和ce是两条高,相交于点m,bf 与cg是两条角平分线,相交于点n,角bmc =100度.
求角bnc的度数.
∵∠BNC=180°-∠NBC-∠NCB
=180°-1/2*∠ABC-1/2*∠ACB
=180°-1/2(∠ABC+∠ACB)
=180°-1/2*(180°-∠A)
=90°+1/2*∠A.
∵∠MBC+∠BCA=90°，
∠NCB+∠ABC=90,°
∴∠MBC+∠MCB+∠ABC+ACB=180°，
∠ABC+∠ACB=180°-（∠MBC+∠MCB)
=∠BMC=100°，
即∠A=180°-（∠ABC+∠ACB)=180°-100°=80°
∴∠BNC=90°+1/2*∠A
=90°+1/2*80°=130°