三角幾何一題 @ zogfly3的部落格

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三角幾何一題

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題目在此: http://ihs.meric.hk/rforum.php/83777.jpg 請解答(需有步驟)

最佳解答:

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

其他解答:

∠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