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由 Raceleader 於 星期日 四月 13, 2003 11:28 am
I will post back the pic later
由 yll 於 星期日 四月 13, 2003 11:01 am
i can't see the question!!!
由 Raceleader 於 星期日 四月 13, 2003 9:26 am
Q1
∠ABG=∠BDG+∠BGD (Exterior angle of triangle)
120°=58°+∠BGD
∠BGD=62°
BG//DE (Given)
∠ABG=∠BDE (Corresponding angles, BG//DE)
∠ABG=∠BDG+∠GDF+∠FDE
120°=58°+∠GDF+24°
∠GDF=38°
CG//DF (Given)
∠CGD=∠GDF=38° (Alternate angles, CG//DF)
BG//DE (Given)
∠BGD=∠GDE (Alternate angles, BG//DE)
38°+v=38°+24°
v=24°
Q2
CD//FE (Given)
∠CDE+∠FED=180° (Interior angles, CD//FE)
∠CBF=180°-∠ABF=180°-k (Adjacent angles on straight line)
∠BFE=180°-∠GFE=180°-90°=90° (Adjacent angles on straight line)
∠BCD+∠CDE+∠DEF+∠EFB+∠FBC=(5-2)180° (Angles sum of polygon)
135°+180°+90°+180°-k=540°
k=45°
∠ABG=∠BDG+∠BGD (Exterior angle of triangle)
120°=58°+∠BGD
∠BGD=62°
BG//DE (Given)
∠ABG=∠BDE (Corresponding angles, BG//DE)
∠ABG=∠BDG+∠GDF+∠FDE
120°=58°+∠GDF+24°
∠GDF=38°
CG//DF (Given)
∠CGD=∠GDF=38° (Alternate angles, CG//DF)
BG//DE (Given)
∠BGD=∠GDE (Alternate angles, BG//DE)
38°+v=38°+24°
v=24°
Q2
CD//FE (Given)
∠CDE+∠FED=180° (Interior angles, CD//FE)
∠CBF=180°-∠ABF=180°-k (Adjacent angles on straight line)
∠BFE=180°-∠GFE=180°-90°=90° (Adjacent angles on straight line)
∠BCD+∠CDE+∠DEF+∠EFB+∠FBC=(5-2)180° (Angles sum of polygon)
135°+180°+90°+180°-k=540°
k=45°
由 Raceleader 於 星期日 四月 13, 2003 9:10 am
They are not difficult...if you are not care about the silly methods
[數學] Question
由 E.T 於 星期日 四月 13, 2003 1:27 am
