如圖ABCD為正方形且B,G,C,E在同一直線上,若角1=角2=角3,AF=5,CG=7,則DG=?(請詳解)?
回答 (2)
2015-12-13 5:25 pm
✔ 最佳答案
Sol在△ADF與△CDE中
AD=CD
∠1=∠3
∠DEC=∠DAF=90度
△ADF全等於△CDE (AAS)
CE=AF=5
∠DFA=180度－∠1－90度＝180度－∠3－90度=∠DEG
∠GDE=∠GDC+∠3=∠GDC+∠2=∠CDF＝AFD=∠DEG
GD=GE=7+5=12
2015-12-13 4:39 am
∠CGD = ∠1 + ∠2 = 2∠1
CD = CG tan ∠CGD = 7 tan(2∠1)
CD = CE / tan∠3 = 5 / tan∠1
7 tan(2∠1) = 5 / tan∠1
14 tan∠1 / (1 - tan²∠1) = 5 / tan∠1
14 tan²∠1 = 5 - 5 tan²∠1
tan²∠1 = 5/19
CD = CG tan(2∠1) = 14 tan²∠1 / (1 - tan²∠1) = 14 (5/19)² / [1 - (5/19)²] = 25/24
DG = √[7² + (25/24)²] = √(28849/576) = (√28849)/24
CD = CG tan ∠CGD = 7 tan(2∠1)
CD = CE / tan∠3 = 5 / tan∠1
7 tan(2∠1) = 5 / tan∠1
14 tan∠1 / (1 - tan²∠1) = 5 / tan∠1
14 tan²∠1 = 5 - 5 tan²∠1
tan²∠1 = 5/19
CD = CG tan(2∠1) = 14 tan²∠1 / (1 - tan²∠1) = 14 (5/19)² / [1 - (5/19)²] = 25/24
DG = √[7² + (25/24)²] = √(28849/576) = (√28849)/24
收錄日期: 2021-04-30 20:22:50
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