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题目 #20137

UUID
e9c09bfc-3642-41a3-8808-411bf0bb9daf
来源
人教版/七下/2026江苏七年级期末数学试卷+答案20.pdf 第 3 页,bbox [31.3, 257.93, 267.87, 329.38]
题型 / 学科
subjective / math / SENIOR
知识树
KT_SENIOR_MATH_STANDARD
难度
难度 2 / 5 简单 · 单一知识点
QA 状态
存疑 / 待修复 缺少答案 公式已降级 重试 0 次,得分 -
最终题 ID
-
错误
处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8091): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8092/v1 -> http://127.0.0.1:8091/v1]

最终题干渲染(f_body 预览)

如图,直线AB,CD相交于点O,OF ⊥ AB,OE 平分 .\(\frac{AOD}{BOD}\)若 = 60,求COE的度数.\(\frac{(1)}{(2)}\)若AOC : COF = 2 :1,求DOE的度数.

答案 / 解析

答案(f_answer)缺少答案

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解析(f_ways)

(1) 平分, = ,fracOEAOE fracAODDOE fracBOD(180 frac6060) =  = −  2 = 60,  = 60,\(\frac{COA}{COE}\) =  + AOE = 60+ 60 = 120 .\(\frac{COA}{COF}\)(2)  :  = 2 :1,\(\frac{AOC}{COF}\)设 = x,则AOC = 2x.   + COF +  = 180,\(\frac{AOC}{2} \frac{FOB}{解得}\) x + x + 90= 180, x = 30,  = 60 .\(\frac{AOC}{平分}\) OE AOD,  = .\(\frac{AOE}{AOE} \frac{DOE}{EOD}\)  +  + AOC = 180, 2 + 60= 180,\(\frac{DOE}{DOE}\) = 60 .

原图与资产(1)

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embedded_image3 [30.0, 344.02, 123.56, 430.97] b79208b061db772b…
打开
/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/e9c09bfc-3642-41a3-8808-411bf0bb9daf/b79208b061db.png

审计记录

阶段审计器结果得分依据(简短)
STRUCTURE structure-rule PASS 0.900
FORMULA formula-auditor FAIL 1.000
KNOWLEDGE knowledge-auditor WARN -
VISION vision-auditor PASS 1.000

知识点(0)

无

事件

时间阶段级别消息
2026-09-24 14:57:44ERRORERROR处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8091): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8092/v1 -> http://127.0.0.1:8091/v1]
2026-09-24 14:53:44FORMULAWARN公式转换未确认,保留原文本: solution
2026-09-24 14:49:43FORMULAWARN公式转换未确认,保留原文本: stem