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弧度BQ與弧度QP分別為32度
2015-06-25 11:17 am
回答 (2)
2015-06-25 12:18 pm
✔ 最佳答案
∠B = ∠Q = ∠D = Arc(AC)/2∠BCD = ∠BCQ + ∠QCD= [Arc(BQ) + Arc(QD)]/2= (32° + 50°)/2= 41° ΔPBC裡面: 外角 = 兩內角和∠P + ∠B= ∠P + ∠Q= ∠BCD= 41°= Answer
2015-06-26 12:47 am
25.
對同弧圓周角為圓心角的一半：
∠BAQ = (1/2)弧BQ = (1/2)(32°) = 16°
∠DCQ = (1/2)弧DQ = (1/2)(50°) = 25°
∠PAQ + ∠BAQ = 180° (直線同側鄰角和)
∠PAQ + 16° = 180°
∠PAQ = 164°
∠PCQ + ∠DCQ = 180° (直線同側鄰角和)
∠PCQ + 25° = 180°
∠PCQ = 155°
∠P + ∠PAQ + ∠Q + ∠PCQ = 360° (四邊形內角和)
∠P + 164° + ∠Q + 155° = 360°
∠P + ∠Q = 41°
2015-06-25 16:48:25 補充：
我回答是因為看不到 ∠B 和 ∠D 在哪裡？
對同弧圓周角為圓心角的一半：
∠BAQ = (1/2)弧BQ = (1/2)(32°) = 16°
∠DCQ = (1/2)弧DQ = (1/2)(50°) = 25°
∠PAQ + ∠BAQ = 180° (直線同側鄰角和)
∠PAQ + 16° = 180°
∠PAQ = 164°
∠PCQ + ∠DCQ = 180° (直線同側鄰角和)
∠PCQ + 25° = 180°
∠PCQ = 155°
∠P + ∠PAQ + ∠Q + ∠PCQ = 360° (四邊形內角和)
∠P + 164° + ∠Q + 155° = 360°
∠P + ∠Q = 41°
2015-06-25 16:48:25 補充：
我回答是因為看不到 ∠B 和 ∠D 在哪裡？
收錄日期: 2021-05-01 13:12:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150625000016KK02963