Trigonometry

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ABCD is a quadrilateral with AB=BC=CD．Angle BAD = 135 degrees
and angle CDA = 75 degrees. Prove that AC=AB
Sol
∠BAC=x
AB=BC
∠BCA=x
∠ABC=180度－2x
在△ABC中依正弦定律
AB/Sinx=AC/Sin(180度－2x)
AB/Sinx=AC/Sin((2x)
AB/AC=Sinx/Sin(2x)=1/(2Cosx)

2011-03-20 23:24:00 補充：
在△ACD中
CD/Sin(135度－x)=AC/Sin75度
AB/Sin(135度－x)=AC/Sin75度
AB/AC=Sin(135度－x)/Sin75度
So
1/(2Cosx)=Sin(135度－x)/Sin75度
2CosxSin(135度－x)=Sin75度
Sin75度=[CosxSin(135度－x)+SinxCos(135度－x)]
+[CosxSin(135度－x)－SinxCos(135度－x)]
Sin75度=Sin135度+Sin(135度－x－x)
Sin(135度－2x)=Sin75度－Sin135度

2011-03-20 23:24:24 補充：
=(√6+√2)/4－√2/2
=(√6－√2)/4
135度－2x=15度
x=60度
△ABC為正△
AC=AB