Question

有高手能帮我解决这个第二小题啊

Answer 1

(1) A=π/6
(2)A=π/6,sinA=1/2,B=5π/6-C
2sinAsinB=1+cosC
sinB=1+cosC,sin(5π/6-C)=1+cosC
(1/2)cosC+(√3/2)sinC=1+cosC
(√3/2)sinC-(1/2)cosC=1
sin(C-π/6)=1,且-π/6<C-π/6<2π/3
C-π/6=π/2
得 C=2π/3,B=π/6
ΔABC中,设AC=m,(m>0)
则BC=m,AB=(√3)m
ΔAEC中,∠AEC=π/6,∠ACE=3π/4
AE=AC·sin∠ACE/sin∠AEC=m·sin(3π/4)/sin(π/6)=(√2)m
ΔABD中,∠ABD=π/6,∠ADB=3π/4
AD=AB·sin∠ABD/sin∠ADB=(√3)m·sin(π/6)/sin(3π/4)=(√6/2)m
所以 λ=AE/AD=(√2)m/((√6/2)m)=(2√3)/3

追问

谢谢(๑• . •๑)