Page 93 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪٨٠‬‬

‫‪S‬‬
‫‪P 5x‬‬

‫‪R 72°‬‬ ‫‪x° Q‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪ :‬ﻟﺪﻳﻨﺎ ‪ .5x = 72 + x‬أي أن ‪ . 4x = 72‬وﻣﻦ ﰒ ﻓﺈن‬
‫‪ . x = 18‬إذن‪.QPS = 5 × 18 = 90° ،‬‬

‫)‪ [Aust.MC 1978] (٤‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ ‪ ABX‬و ‪ ACY‬ﻣﺴﺘﻘﻴﻤﺎن‪ .‬ﻣﻨﺼﻔﺎ‬

‫اﻟﺰاوﻳﺘﲔ ‪ XBC‬و ‪ BCY‬ﻳﻠﺘﻘﻴﺎن ﰲ اﻟﻨﻘﻄﺔ ‪ .BZC = 80° .Z‬ﻣﺎ ﻗﻴﺎس‬

‫‪ BAC‬؟‬

‫)د( ‪35°‬‬ ‫)ج( ‪30°‬‬ ‫)ب( ‪25°‬‬ ‫)أ( ‪20°‬‬

‫‪X‬‬

‫‪θ°‬‬ ‫‪B‬‬ ‫‪Y‬‬
‫‪A‬‬ ‫‪α°‬‬

‫‪α°‬‬
‫‪Z‬‬

‫‪β°‬‬
‫‪β°‬‬

‫‪C‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )أ(‪ :‬ﰲ ‪ △ABC‬ﻟﺪﻳﻨﺎ‬

‫‪θ + (180 − 2α) + (180 − 2β) = 180 .‬‬

‫إذن‪ . θ = 2(α + β) − 180 ،‬وﻣﻦ زواﻳﺎ اﳌﺜﻠﺚ ‪ △BCZ‬ﳒﺪ أن‬
‫‪ . α + β + 80 = 180‬أي أن ‪ . α + β = 100‬إذن‪،‬‬
‫‪. θ = 2(α + β) − 180 = 2 ×100 − 180 = 20°‬‬