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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
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？
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
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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？
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