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標題:
我想要中一的幾何題,40分!
發問:
我想要中一的幾何題,程度要愈難愈好 題目只可以用到以下的定律: angles at a point adj. angles on st. line adj. angles supp. vert. opp angles corr. angles, AB//CD alt. angles, AB//CD int. angles, AB//CD corr. angles equal alt. angles equal int. angles supp. ext. angle of triangle angle sum of triangle
最佳解答:
圖片參考：http://i169.photobucket.com/albums/u209/lautszki/F.jpg Find the following angles in only one step: i) ∠AOB ii) ∠AOD iii) ∠BOE iv) ∠AED v) ∠BDE vi) ∠BOC vii) ∠BOF viii) ∠AOF ix) ∠COD x) Find∠DOE by using the result of∠AOB xi) Find∠FOE by using the result of ∠BOF and ∠DOE. xii) By using the result of∠AOB, ∠AOD, ∠BOF and ∠DOE, find∠FOH if ∠HOE=20°. (adj.∠s on st. line is NOT allowed to use) Solution: i) ∠AOB = 180°-55°-60° (∠sum of△) = 65° ii) ∠AOD = 55°+60° (ext.∠of△) = 115° iii) ∠BOE = 55°+60° (ext.∠of△) = 115° iv) ∠AED = ∠BAO (alt.∠s, BA//ED) = 60° v) ∠BDE = ∠ABO (alt.∠s, BA//ED) = 55° vi) ∠BOC = 180°-55° (int.∠s, BA//CF) = 125° vii) ∠BOF = ∠ABO (alt.∠s, BA//CF) = 55° viii) ∠AOF = 180°-60° (int.∠s, BA//CF) = 120° ix) ∠COD =∠ABO (corr.∠s, BA//CF) = 55° x) ∠DOE = ∠AOB(vert. opp.∠s) = 65° (the result from i) xi) ∠FOE = 180°-∠BOF-∠DOE (adj.∠s on st. line) = 180°- 55° - 65° = 60° xii) ∠FOH = 360°–∠AOB –∠AOD –∠BOF –∠DOE –∠HOE (∠s at a pt.) = 360°– 65°–115°–55°–65°–20° = 40°
你好！ 我建議你花幾十元到書局親自選購一本補充練習。 在商務書局見過一本橙色、英文的數學補充，本子小小但題目齊全，可惜被媽媽弄掉了，不能提供相關書名、出版社。有時間可到各大書局找找適合自己的。|||||打中文啦!!
我想要中一的幾何題,40分!
發問:
我想要中一的幾何題,程度要愈難愈好 題目只可以用到以下的定律: angles at a point adj. angles on st. line adj. angles supp. vert. opp angles corr. angles, AB//CD alt. angles, AB//CD int. angles, AB//CD corr. angles equal alt. angles equal int. angles supp. ext. angle of triangle angle sum of triangle
最佳解答:
圖片參考：http://i169.photobucket.com/albums/u209/lautszki/F.jpg Find the following angles in only one step: i) ∠AOB ii) ∠AOD iii) ∠BOE iv) ∠AED v) ∠BDE vi) ∠BOC vii) ∠BOF viii) ∠AOF ix) ∠COD x) Find∠DOE by using the result of∠AOB xi) Find∠FOE by using the result of ∠BOF and ∠DOE. xii) By using the result of∠AOB, ∠AOD, ∠BOF and ∠DOE, find∠FOH if ∠HOE=20°. (adj.∠s on st. line is NOT allowed to use) Solution: i) ∠AOB = 180°-55°-60° (∠sum of△) = 65° ii) ∠AOD = 55°+60° (ext.∠of△) = 115° iii) ∠BOE = 55°+60° (ext.∠of△) = 115° iv) ∠AED = ∠BAO (alt.∠s, BA//ED) = 60° v) ∠BDE = ∠ABO (alt.∠s, BA//ED) = 55° vi) ∠BOC = 180°-55° (int.∠s, BA//CF) = 125° vii) ∠BOF = ∠ABO (alt.∠s, BA//CF) = 55° viii) ∠AOF = 180°-60° (int.∠s, BA//CF) = 120° ix) ∠COD =∠ABO (corr.∠s, BA//CF) = 55° x) ∠DOE = ∠AOB(vert. opp.∠s) = 65° (the result from i) xi) ∠FOE = 180°-∠BOF-∠DOE (adj.∠s on st. line) = 180°- 55° - 65° = 60° xii) ∠FOH = 360°–∠AOB –∠AOD –∠BOF –∠DOE –∠HOE (∠s at a pt.) = 360°– 65°–115°–55°–65°–20° = 40°
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其他解答:你好！ 我建議你花幾十元到書局親自選購一本補充練習。 在商務書局見過一本橙色、英文的數學補充，本子小小但題目齊全，可惜被媽媽弄掉了，不能提供相關書名、出版社。有時間可到各大書局找找適合自己的。|||||打中文啦!!
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