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第一種情形：
 檢視圖片 
tan a = √3/(k-1)，tan b = [(3√3)/2]/[k-(3/2)] = 3√3/(2k-3)
tan (a+b) = tan 30度
(tan a + tan b)/[1-(tan a*tan b)] = 1/√3
[√3/(k-1) + 3√3/(2k-3)]/{1-[√3/(k-1)][3√3/(2k-3)]} = 1/√3
上式左邊分子分母同乘以(k-1)(2k-3)
[√3(2k-3)+3√3(k-1)]/[(k-1)(2k-3)-9] = 1/√3
交叉相乘得
3(2k-3)+9(k-1) = (k-1)(2k-3)-9
乘開整理得
k^2 - 10k + 6 = 0
k = 5±√19 (負不合)

第二種情形：

檢視圖片 

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