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用余弦定理c²=a²+b²-2abcosC求出c=2
再用余弦定理求出cosA=1/2则A=60°则B=180°-60°-45°=75° a=√6, b=√3+1, C=45°
c^2 = a^2+b^2 -2abcosC
= 6+(4+2√3) - 2√6(√3+1)(√2/2)
= 10+2√3- (6+2√3)
= 4
c =2
a/sinA = c/sinC
√6/sinA = 2/sin45°
√6/sinA = 2√2
=> sinA = √3/2
A= 60°
B = 180°-60°-45°=75°
再用余弦定理求出cosA=1/2则A=60°则B=180°-60°-45°=75° a=√6, b=√3+1, C=45°
c^2 = a^2+b^2 -2abcosC
= 6+(4+2√3) - 2√6(√3+1)(√2/2)
= 10+2√3- (6+2√3)
= 4
c =2
a/sinA = c/sinC
√6/sinA = 2/sin45°
√6/sinA = 2√2
=> sinA = √3/2
A= 60°
B = 180°-60°-45°=75°
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