【题目】四边形ABCD中,BC⊥CD,∠BCA=600,∠CDA=1350,。求AD边的长.
【答案】8
【解析】
作AF⊥AC于F,作AE⊥CD交CD的延长线于E,利用三角形的面积公式S△ABC=BCAF求得AF的长,在△AFC中,根据∠BCA的正弦求得AC的长,再证明△ADE为等腰直角三角形,由此即可求解.
作AF⊥AC于F,作AE⊥CD交CD的延长线于E。可求AC=16,AD=8.
作AF⊥BC于F,作AE⊥CD交CD的延长线于E.
∵S△ABC=BCAF=×10×AF=40,
∴AF=8,
∵sin∠BCA=sin60°=AF:AC=,
∴AC=16.
∵BC⊥CD,AE⊥CD
∴∠CAE=∠BCA=60°,
∴∠ACD=90°-60°=30°,
∵∠CDA=135°,
∴AE=ED=sin∠ACDAC=8.
在等腰直角三角形中AD=AE=8.
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