三角幾何一題

From F E A B

X = Y + 30 ---------- 1

From C D E F

4X + Y = 180 ---------------- 2

Put 1 into 2

4 ( Y + 30 ) + Y = 180

4Y + 120 + Y = 180
5Y = 60
Y = 12

PUT ( Y = 12 ) INTO 1

X = Y + 30
X = 42

2007-04-10 09:22:09 補充：
X = 30 + Y (alt. anlges of parallel lines AB & EF) --------------------- 14X + Y = 180 (int. angles of parallel lines CD & EF) ----------------- 2

∠4x + ∠y = 180∘(int.∠s, //lines)------------1)
∠x = ∠y+30∘(alt.∠s, //lines) -------------2)
From 2), x=y+30
y=x-30 ......3)
Substitute 3) to 1),
4x+x-30=180
5x=210
x=42
∵ x=42 and y=x-30
∴ y=12 , x= 42

easy, first of all, we need to have all the equations to be solved.

4x + y = 180 (int. angles of parallel lines CD & EF) ----------------- (1)
x = 30 + y (alt. anlges of parallel lines AB & EF) --------------------- (2)

substitute equation (2) into equation (1)

4(30 + y) + y = 180
120 + 5y = 180
5y = 60
y = 30 degrees
x = 60 degrees

4x + y = 180
x = y + 30

4(y+30) + y = 180
4y + 120 + y = 180
5y = 60

y=12, x = 42