Question

已知:如图,AD,BE分别是∠BAC和∠ABC的角平分线.且∠BAD=41°36′12″，∠ABC=75°12′。求：

已知:如图,AD,BE分别是∠BAC和∠ABC的角平分线.且∠BAD=41°36′12″，∠ABC=75°12′。求：∠DAC，∠BAC，∠EBC，∠ABE　的度数。
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要因为和所以 ∵ ∴~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~·

Answer 1

∵AD,BE分别是∠BAC和∠ABC的角平分线
∴∠DAC=∠BAD
∠BAC=2*∠DAC
所以∠DAC=∠BAD=41°36′12″
又因为∠BAC=2*∠DAC=2*41°36′12″=83°12′24″
∠EBC=∠ABE=1/2*∠ABC=75°12′=37°36′
再没了

Answer 2

解：∠DAC=∠BAD=41°36′12″
∠BAC=2∠BAD=83°12′24″
∠EBC=∠ABC/2=37°36′
∠ABE=∠EBC=37°36′

Answer 3

∵AD,BE分别是∠BAC和∠ABC的角平分线
∴∠DAC=∠BAD
∠BAC=2*∠DAC
∠DAC=∠BAD=41°36′12″
∠BAC=2*∠DAC=2*41°36′12″=83°12′24″
∠EBC=∠ABE=1/2*∠ABC=75°12′=37°36′