在Rt三角形ABC中,AB=AC=2,∠A=90°,现取一块等腰直角三角板,将45°角的顶点放在斜边BC的中点O处,三角板的直角边与线段AB、AC分别交于点E、F,设BE=x,CF=y,∠BOE=a(45°≤a≤90°)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/27 16:37:33
![在Rt三角形ABC中,AB=AC=2,∠A=90°,现取一块等腰直角三角板,将45°角的顶点放在斜边BC的中点O处,三角板的直角边与线段AB、AC分别交于点E、F,设BE=x,CF=y,∠BOE=a(45°≤a≤90°)](/uploads/image/z/6689894-14-4.jpg?t=%E5%9C%A8Rt%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAB%3DAC%3D2%2C%E2%88%A0A%3D90%C2%B0%2C%E7%8E%B0%E5%8F%96%E4%B8%80%E5%9D%97%E7%AD%89%E8%85%B0%E7%9B%B4%E8%A7%92%E4%B8%89%E8%A7%92%E6%9D%BF%2C%E5%B0%8645%C2%B0%E8%A7%92%E7%9A%84%E9%A1%B6%E7%82%B9%E6%94%BE%E5%9C%A8%E6%96%9C%E8%BE%B9BC%E7%9A%84%E4%B8%AD%E7%82%B9O%E5%A4%84%2C%E4%B8%89%E8%A7%92%E6%9D%BF%E7%9A%84%E7%9B%B4%E8%A7%92%E8%BE%B9%E4%B8%8E%E7%BA%BF%E6%AE%B5AB%E3%80%81AC%E5%88%86%E5%88%AB%E4%BA%A4%E4%BA%8E%E7%82%B9E%E3%80%81F%2C%E8%AE%BEBE%3Dx%2CCF%3Dy%2C%E2%88%A0BOE%3Da%EF%BC%8845%C2%B0%E2%89%A4a%E2%89%A490%C2%B0%EF%BC%89)
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