Page 47 - Engineering
P. 47
ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ( ٣٤
AB
50°
Cx
100°
DE
اﻟﺤﻞ :اﻹﺟﺎﺑﺔ ﻫﻲ )د( :ارﺳﻢ ﻣﺴﺘﻘﻴﻤﺎً ﻣﻮازﻳﺎً ﻟﻠﻤﺴﺘﻘﻴﻢ DEوﳝﺮ ﺑﺎﻟﻨﻘﻄﺔ C
وﻟﻴﻜﻦ .CF
AB
CF
DE
ﻋﻨﺪﺋﺬ FCD = 180° − 100° = 80° ،و FCB = 50°ﺑﺎﻟﺘﺒﺎدل اﻟﺪاﺧﻠﻲ.
إذن.xɵ = FCD + FCB = 80° + 50° = 130° ،
) (٢٣ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ ،ﻗﻴﻤﺔ x − yﺗﺴﺎوي
)د( 40° )ج( 30° )ب( 20° )أ( 10°
D x° C 120°
y°
E 80°
BA
اﻟﺤﻞ :اﻹﺟﺎﺑﺔ ﻫﻲ )ب( :ﻟﺪﻳﻨﺎ y + 80 = 120°ﺑﺎﻟﺘﺒﺎدل اﻟﺪاﺧﻠﻲ .إذن،
