標題:

請教!中四圓形

發問:

http://i296.photobucket.com/albums/mm161/april_16_02_2008/DSC00024.jpg1.)圖中,PQ是圓於A的切線,且DB平行於PQ.求x http://i296.photobucket.com/albums/mm161/april_16_02_2008/DSC00025.jpg 2.)圖中,求∠CBD

免費註冊體驗

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

Q.1 Since DB//PQ, therefore, angle DBA = angle BAQ = 55 (alternate angles DB//PQ) Angle BDA = angle BAQ = 55 (angles in alternate segment) Therefore, angle BDA = angle DBA = 55. Therefore, angle DAB = 180 - 55 - 55 =70 ( angle sum of triangle) Therefore, x = 180 - 70 = 110 (opposite angle of cyclic quad.) 2008-06-05 17:18:19 補充: Q.2 Angle ABC = 360 - 40 - 30 - 75 - 75 - 35 = 105.Therefore, angle ABC = angle ADE = 105.Therefore, ABCD is a cyclic quadrilateral (ext. angle equals int. opposite angle).Therefore, angle CBD = angle CAD = 75 (angle in the same segment.)

其他解答:

第一題我上唔到去 我識第2題 求∠CBD ∠D=180度-105度 ∠D=75度 180度=∠A+30度+75度 ∠A=75度 ∠CBD=360度-70度-35度-75度-75度=105度|||||1.有問題 夾角無理由係同圓周嫁喎~ 2.∠CDA=180°-105°=75°- ∠CAD=180°-30°-75°=75° ∠CBD=360°-70°-35°-75°-75°=105°|||||小朋友,點解你隻∠55度會掂到個circle嫁?
文章標籤
創作者介紹

ccf83dy60d的部落格

ccf83dy60d 發表在 痞客邦 PIXNET 留言(0) 人氣()