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Mathematical reasoning in junior high school
⑴∠AOC =∠AOB+∠BOC = 90+60 = 150

∫OM is the bisector of ∞∠AOC.

∴∠AOM=( 1/2)∠AOC=75

∴∠MOB=∠AOB-∠AOM=90 -75 = 15

∵ON is the bisector of∝∠ ∝∠BOC.

∴∠BON=( 1/2)∠BOC=30

∴∠MON=∠MOB+∠BON=45

⑵∠MON=α/2

(3) ∠ mon is related to α and has nothing to do with β.

∠∠AOC =∠AOB+∠BOC =α+β

∫OM is the bisector of ∞∠AOC.

∴∠aom=( 1/2)∠aoc=(α/2)+(β/2)

∴∠mob=∠aob-∠aom=α-[(α/2)+(β/2)]=(α/2)-(β/2)

∵ON is the bisector of∝∠ ∝∠BOC.

∴∠BON=( 1/2)∠BOC=β/2

∴∠mon=∠mob+∠bon=(α/2)-(β/2)+(β/2)=α/2