maths

AB = BC = AC
Hence, ΔABC is an equilateral triange.
∠ABD = 60°

Let y = BD

In ΔABD :
tan∠ABD = AD/BD
tan60° = AD/h
√3 = AD/h
AD = h√3

Since AD = AE + ED and AE = ED
then AD = 2ED
h√3 = 2ED
ED = (h√3)/2

In ΔBED :
tan∠EBD = ED/BD
tan∠EBD = [(h√3)/2]/h
tan∠EBD = (√3)/2
∠EBD = 40.89°

θ = 60° - 40.89°
θ = 19.11°

is AD⊥BC ? this is EXTREMELY IMPORTANT !!!!
if yes, then :

let AB =BC =CA = a

BD = DC = a/2, (prop. of equil.Δ)
AD² = a² - (a/2)²
AD = (√3)a/2
AE = ED = (√3)a/4

BE = √((a/2)²+((√3)a/4)²) = (√7)a/4
∠BAD = 30' (prop. of equil.Δ)

AE/sinΘ = BE /sin∠BAD
((√3)a/4) / sinΘ = ((√7)a/4)/(0.5)
√3 / sinΘ =c
sin Θ = (√3) / (2√7)

if exact value is note required, you can use answer 001 to find the value of Θ first, then calculate sinΘ, which is much quicker than my method.

2012-05-25 19:51:41 補充：
last two steps should be:

√3 / sinΘ = 2√7
sin Θ = (√3) / (2√7)