Interesting Trigonometry 2009

ΔABC:
20o + (10o + 70o) + (∠ABD + 20o) = 180o (Δ內角和)
∠ABD = 60o
∠CAB = ∠CBA = 90o
所以 ΔABC 是等腰三角形，CA = CB
設 CA = CB = y

ΔBDC:
∠BDC = (10 + 70o) + 60o = 140o (ΔABD外角)
DC/sin20o = CB/sin∠BDC (正弦定律)
DC/sin20o = y/sin140o
DC = 0.5321y

ΔACE:
∠AEC = 70o + (60o + 20o) = 150o (ΔABE外角)
CE/sin10o = CA/sin∠AEC (正弦定律)
CE/sin10o = y/sin150o
CE = 0.3473y

ΔCDE:
DE2 = DC2 + CE2 - 2DC.CE.cos20o (餘弦定律)
DE2 = (0.5321y)2 + (0.3473y)2 - 2(0.5321y)(0.3473)cos20o
DE = 0.2376y

ΔADE:
AD = CA - DC = y - 0.5321y = 0.4679y
AD/sinx = DE/sin10o (正弦定律)
0.4679y/sinx = 0.2376y/sin10o
x = 20o

「0.4679y/sinx = 0.2376y/sin10度 => x = 20度」
利用計算機，恐怕只能說是近似值，有點遺憾。

這題我曾在年半前在論壇內看過.