※ 引述《notecook (邦邦)》之銘言：
: 三角形ABC中a^2-c^2=bc，∠B=51度，試求∠C=？
a^2 - c^2 = bc
=> cosA = (b^2+c^2-a^2)/(2bc) = (b-c)/(2c) = (sinB-sinC)/(2sinC)
= -bc
cosC = (b^2+a^2-c^2)/(2ba) = (b+c)/(2a) = (sinB+sinC)/(2sinA)
= bc
=> sin(A-C) = sinAcosC - cosAsinC = (sinB+sinC)/2 - (sinB-sinC)/2 = sinC
=> ∠A = 2∠C
∠C = (180°- 51°)/3 = 43°
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