Page 110 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪٩٧‬‬

‫)د( ‪106°‬‬ ‫)ج( ‪96°‬‬ ‫)ب( ‪94°‬‬ ‫)أ( ‪74°‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )د(‪ :‬ﻻﺣﻆ أن ‪) 116° = 42° + B‬ﺧﺎرﺟﺔ ﻋﻦ اﳌﺜﻠﺚ‬

‫‪ .(ABC‬إذن‪ .B = 74° ،‬و‪‬ﺬا ﻓﺈن ‪) ɵy = B = 74°‬ﺑﺎﻟﺘﻨﺎﻇﺮ(‪.‬‬

‫و‪‬ﺬا ﻳﻜﻮن ‪) xɵ = 180° − ɵy = 180° − 74° = 106°‬زاوﻳﺔ ﻣﺴﺘﻘﻴﻤﺔ(‪.‬‬

‫‪، AED = 90°‬‬ ‫)‪ (٢٧‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪،BAC = 36° ،AE BCD ،‬‬
‫‪ .AB = AC = CD‬ﻣﺎ ﻗﻴﻤﺔ ‪ x‬؟‬

‫‪AE‬‬

‫‪36°‬‬

‫‪x‬‬

‫‪BC‬‬ ‫‪D‬‬

‫)د( ‪72°‬‬ ‫)ج( ‪67°‬‬ ‫)ب( ‪54°‬‬ ‫)أ( ‪36°‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪ B = BCA = 72° :‬ﻷن ﳎﻤﻮع زواﻳﺎ اﳌﺜﻠﺚ ‪ABC‬‬

‫ﻳﺴﺎوي ‪ 180°‬وأن ‪ .AB = AC‬وﻟﺬا ﻓﺈن‬

‫‪) ACD = 36° + 72° = 108°‬ﺧﺎرﺟﺔ ﻋﻦ اﳌﺜﻠﺚ ‪.(ABC‬‬

‫‪ CAD = CDA = 36°‬ﻷن ﳎﻤﻮع زواﻳﺎ اﳌﺜﻠﺚ ‪ ACD‬ﻳﺴﺎوي ‪ 180°‬وأن‬

‫‪ .AC = CD‬أﻳﻀﺎً‪،‬‬
‫‪ EAD = CDA = 36°‬ﺑﺎﻟﺘﺒﺎدل‪.‬‬

‫إذن‪.x = 180° − (36° + 90°) = 54° ،‬‬