Page 142 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١٢٩‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪ :‬ﻟﻨﻔﺮض أن ‪ .ABP = x‬ﻋﻨﺪﺋﺬ‪،‬‬

‫‪ABP = QBP = QCB = QCP = x‬‬

‫إذن‪) 40 + 2x + 2x = 180° ،‬ﳎﻤﻮع زواﻳﺎ ‪ .(△ABC‬ﻣﻦ ذﻟﻚ ﻧﺮى أن‬
‫‪ .x = 35°‬اﻵن‪) 40 + 35° + APB = 180° ،‬ﳎﻤﻮع زواﻳﺎ ‪ .(△APB‬إذن‪،‬‬

‫‪. APB = 105°‬‬

‫)‪ (٦٩‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ .QR AC .QR = 10 ،PQ = 2 ،AB = 30 ،‬ﻣﺎ‬
‫ﻃﻮل ‪ BC‬ﻷﻗﺮب ﻋﺪد ﺻﺤﻴﺢ ؟‬

‫‪QB‬‬
‫‪R‬‬

‫‪P‬‬
‫‪AC‬‬

‫)أ( ‪) 4‬ب( ‪) 5‬ج( ‪) 6‬د( ‪7‬‬
‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ج(‪ :‬ﲟﺎ أن ‪ QR AC‬ﻓﺈن ‪ QRP = BAC‬ﺑﺎﻟﺘﺒﺎدل‪ .‬وﻟﺬا‬

‫ﻓﺈن ‪ .(AA) △ABC ∼ △RPQ‬إذن‪،‬‬
‫‪ . BC = PQ‬وﻟﻜﻦ‬

‫‪AB RP‬‬

‫‪PR = (QP )2 + (QR )2 = 4 + 100 = 104‬‬
‫إذن‪.BC = PQ × AB = 2 × 30 = 5.88 ،‬‬

‫‪RP 104‬‬

‫و‪‬ﺬا ﳒﺪ أن ‪) BC = 6‬ﻷﻗﺮب ﻋﺪد ﺻﺤﻴﺢ(‪.‬‬

‫)‪ (٧٠‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ △ACB ،‬ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﻋﻨﺪ ‪،AF = FB ،C‬‬