高中数学解三角形大题，老师帮忙在
解：(1)∵m//n
∴1/(sin(B+C)/2)² = 4/(cos2A+7/2)
4(sin(B+C)/2)² =cos2A+7/2
4(cosA/2)²= cos2A+7/2
2(cosA+1) = 2(cosA)² - 1+ 7/2
4(cosA)²-4cosA+1 =0
(2cosA-1)² =0
得：cosA = 1/2
A= π/3
(2)a=√3，
b+c=3
c = 3-b
由正弦定理：a/sinA = b/sinB = c/sinC
√3/sinπ/3 = b/sinB = (3-b)/sin(π-A-B)
...全部
解：(1)∵m//n
∴1/(sin(B+C)/2)² = 4/(cos2A+7/2)
4(sin(B+C)/2)² =cos2A+7/2
4(cosA/2)²= cos2A+7/2
2(cosA+1) = 2(cosA)² - 1+ 7/2
4(cosA)²-4cosA+1 =0
(2cosA-1)² =0
得：cosA = 1/2
A= π/3
(2)a=√3，
b+c=3
c = 3-b
由正弦定理：a/sinA = b/sinB = c/sinC
√3/sinπ/3 = b/sinB = (3-b)/sin(π-A-B)
2 = b/sinB = (3-b)/sin(2π/3-B)
2 = b/sinB
sinB = b/2
cosB = √(4-b^2)/2
2 = (3-b)/sin(2π/3-B)
2sin(2π/3-B) = 3-b
2(√3/2cosB+1/2sinB) = 3-b
√3cosB+sinB = 3-b
√3(√(4-b^2)/2) + b/2 = 3-b
√(12-3b^2) = 6-3b
12-3b^2 = 36 -36b+9b^2
b^2-3b+2 =0
(b-1)(b-2) =0
b1 =1，b2=2(舍去）
b =1 ，c=2
△ABC的面积 = 1/2bc sinA
= 1/2×2×1×(√3/2)
= √3/2
sinB = b/2= 1/2
B = π/6。
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