解答 问题数学? 5
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既然 C = (π/3 + B),那么:
sinC = sin(π/3 + B)
= sin(π/3)cosB + cos(π/3)sinB
= √3/2 * cosB + 1/2 * sinB = 2sinB
√3cosB + sinB = 4sinB
3sinB = √3 cosB
tanB = sinB/cosB = √3/3
所以,B = π/6
那么:C = π/3 + B = π/2
可见,△ABC 是一个直角三角形
根据正弦定理的性质,有外接圆半径 R:
2R = a/sinA = b/sinB = c/sinC = 8√3
所以,a = 2RsinA = 8√3 * sin(C-B) = 8√3 * √3/2 = 12
b = 2RsinB = 8√3 * 1/2 = 4√3
所以,S△ABC = 1/2 * a * b = 1/2 * 12 * 4√3 = 24√3
sinC = sin(π/3 + B)
= sin(π/3)cosB + cos(π/3)sinB
= √3/2 * cosB + 1/2 * sinB = 2sinB
√3cosB + sinB = 4sinB
3sinB = √3 cosB
tanB = sinB/cosB = √3/3
所以,B = π/6
那么:C = π/3 + B = π/2
可见,△ABC 是一个直角三角形
根据正弦定理的性质,有外接圆半径 R:
2R = a/sinA = b/sinB = c/sinC = 8√3
所以,a = 2RsinA = 8√3 * sin(C-B) = 8√3 * √3/2 = 12
b = 2RsinB = 8√3 * 1/2 = 4√3
所以,S△ABC = 1/2 * a * b = 1/2 * 12 * 4√3 = 24√3
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