Question

初一数学题做不来，求解两道，急啊！如图：∠ABC=50°.∠ACB=60°，∠ABC,∠ACB的平分

初一数学题做不来，求解两道，急啊！如图：∠ABC=50°.∠ACB=60°，∠ABC,∠ACB的平分线交于点o过点o作EF／／BC交于AB于E，交AC于F，求∠BOC的度数。

Answer 1

∵EF∥BC
∴∠AEF=∠ABC,∠AFE=∠ACB
∵OB平分∠ABC,OC平分∠ACB
∴∠EBO=∠CBO,∠CFO=∠BCO
∠BOC=180°-∠OBC-∠OCB

=180°-(1/2)(∠ABC+∠ACB)
=125°

Answer 2

∵EF∥BC∴∠BOE=∠OBC=1／2∠ABC=25°
∠COF=∠OCB=1／2∠ACB=30°
∠BOC=180°-∠BOE-∠COF
=180°-25°-30°
=125°

Answer 3

∵EF∥BC
∴∠AEF=∠ABC,∠AFE=∠ACB
∵OB平分∠ABC,OC平分∠ACB
∴∠EBO=∠CBO,∠CFO=∠BCO
∠BOC=180°-∠OBC-∠OCB

=180°-(1/2)(∠ABC+∠ACB)
=125°

Answer 4

由题目知道∠ABC=50°.∠ACB=60°，∠ABC,∠ACB的平分线交于点O，则有：∠EBO=∠OBC=25°

∠FCO=∠OCB=30°，三角形的内角和为180°，则∠BOC=180°-∠OBC-∠OCB=180°-25°-30°=125°

Answer 5

∵EF∥BC
∴∠AEF=∠ABC,∠AFE=∠ACB
∵OB平分∠ABC,OC平分∠ACB
∴∠EBO=∠CBO,∠CFO=∠BCO
∠BOC=180°-∠OBC-∠OCB

=180°-(1/2)(∠ABC+∠ACB)
=125°

Answer 6

∠BOC=180°-∠ABC/2-∠ACB/2=125°

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