<後日交`40 marks> F.2 三角比 (1)

Assume it is BC = CD = DE = EF and not BC//CD//DE//EF.
Let AF = b, and BC = a.
So tan 10 = b/4a, b/a = 4 tan 10 = 0.70530792
Angle ACF = 10 + t ( t means theta) (ext. angle of triangle).
so tan ( 10 + t ) = AF/CF = b/3a = 0.70530792/3 = 0.23510264
so 10 + t = arctan (0.23510264) = 13.23, so t = theta = 13.23 - 10 = 3.23.
tan (angle ADF) = b/2a = 0.3526539, so angle ADF = 19.425
tan (angle AEF) = b/a = 0.70530792, so angle AEF = 35.195
so angle DAE = phi = angle AEF - angle ADF = 35.195 - 19.425
= 15.77 ( ext. angle of triangle).

Re STEVIE-G™ :

肯定冇,
因為 WS 比既資料就系咁多。
其實仲有冇其他 method 可以計到呢 ?

條題目你有冇打漏條件...因為得呢D應該搵唔到個角度...