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標題:
Interesting Trigonometry 2009
發問:
As follows: 圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/CCF27072009_00000.jpg (Link: http://i707.photobucket.com/albums/ww74/stevieg90/CCF27072009_00000.jpg)
最佳解答:
ΔABC: 20o + (10o + 70o) + (∠ABD + 20o) = 180o (Δ內角和) ∠ABD = 60o ∠CAB = ∠CBA = 90o 所以 ΔABC 是等腰三角形，CA = CB 設 CA = CB = y ΔBDC: ∠BDC = (10 + 70o) + 60o = 140o (ΔABD外角) DC/sin20o = CB/sin∠BDC (正弦定律) DC/sin20o = y/sin140o DC = 0.5321y ΔACE: ∠AEC = 70o + (60o + 20o) = 150o (ΔABE外角) CE/sin10o = CA/sin∠AEC (正弦定律) CE/sin10o = y/sin150o CE = 0.3473y ΔCDE: DE2 = DC2 + CE2 - 2DC.CE.cos20o (餘弦定律) DE2 = (0.5321y)2 + (0.3473y)2 - 2(0.5321y)(0.3473)cos20o DE = 0.2376y ΔADE: AD = CA - DC = y - 0.5321y = 0.4679y AD/sinx = DE/sin10o (正弦定律) 0.4679y/sinx = 0.2376y/sin10o x = 20o
「0.4679y/sinx = 0.2376y/sin10度 => x = 20度」 利用計算機，恐怕只能說是近似值，有點遺憾。|||||這題我曾在年半前在論壇內看過.
Interesting Trigonometry 2009
發問:
As follows: 圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/CCF27072009_00000.jpg (Link: http://i707.photobucket.com/albums/ww74/stevieg90/CCF27072009_00000.jpg)
最佳解答:
ΔABC: 20o + (10o + 70o) + (∠ABD + 20o) = 180o (Δ內角和) ∠ABD = 60o ∠CAB = ∠CBA = 90o 所以 ΔABC 是等腰三角形，CA = CB 設 CA = CB = y ΔBDC: ∠BDC = (10 + 70o) + 60o = 140o (ΔABD外角) DC/sin20o = CB/sin∠BDC (正弦定律) DC/sin20o = y/sin140o DC = 0.5321y ΔACE: ∠AEC = 70o + (60o + 20o) = 150o (ΔABE外角) CE/sin10o = CA/sin∠AEC (正弦定律) CE/sin10o = y/sin150o CE = 0.3473y ΔCDE: DE2 = DC2 + CE2 - 2DC.CE.cos20o (餘弦定律) DE2 = (0.5321y)2 + (0.3473y)2 - 2(0.5321y)(0.3473)cos20o DE = 0.2376y ΔADE: AD = CA - DC = y - 0.5321y = 0.4679y AD/sinx = DE/sin10o (正弦定律) 0.4679y/sinx = 0.2376y/sin10o x = 20o
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其他解答:「0.4679y/sinx = 0.2376y/sin10度 => x = 20度」 利用計算機，恐怕只能說是近似值，有點遺憾。|||||這題我曾在年半前在論壇內看過.
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