中二數學問題！(急) 10 分

Let AF = h
Let BC = CD = DE = EF = a

Consider ΔABF:
tan10o = h/(4a)
Hence, h = 4a tan10o ...... (*)

Consider ΔACF:
tan∠ACF = h/(3a)
tan∠ACF = (4a tan10o)/3a
∠ACF = 13.2o
Since∠ACF = θ + 10o (ext. ∠ of ΔABC)
thus θ + 10o = 13.2o
θ = 3.2o

Consider ΔADF:
tan∠ADF = h/2a
tan∠ADF = (4a tan10o)/(2a)
∠ADF = 19.4o

Consider ΔAEF:
tan∠AEF = h/a
tan∠AEF = (4a tan10o)/a
∠AEF = 35.2o

∠ADF + φ = ∠AEF (ext. ∠ of ΔADE)
19.4o + φ = 35.2o
φ = 15.8o

Ans: θ = 3.2o and φ = 15.8o
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θ = 3.2o and φ = 15.8o