MATHS (Triagular problem)

The length of a side of the triangle is sqrt(21).

2013-12-30 21:22:14 補充：
Let a be the angle DCB, so angle DCA is (60 - a).Let x be the length of the side of the triangle, by using Cosine Formula,BD^2 = BC^2 + CD^2 - 2(BC)(CD)cos a==> 3 = x^2 + 9 - 6x cos a==> cos a = (x^2 + 6) / (6x)therefore, sin a = √[(6x)^2 - (x^2 + 6)^2] / (6x)= √(24x^2 - x^4 - 36) / (6x)Also, AD^2 = AC^2 + CD^2 - 2(AC)(CD)cos (60 - a)==> 12 = x^2 + 9 - 6x(cos 60 cos a + sin 60 sin a)==> 2x^2 - 6 = 6x cos a + √3 (6x sin a)==> 2x^2 - 6 = x^2 + 6 + √3(24x^2 - x^4 - 36)==> √(72x^2 - 3x^4 - 108 = x^2 - 12==> 72x^2 - 3x^4 - 108 = x^4 - 24x^2 + 144==> 4x^4 - 96x^2 + 252 = 0==> 4(x^2 - 3)(x^2 - 21) = 0==> x = √21 or -√21 (rej) or √3 (rej, too small) or -√3 (rej)
Ans : The length of the side of the triangle is √21 units.

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那些年可否列出步驟？