boss43
春芽
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根据正弦定理,有
BC/sinA = AB/sinC,则有BC = AB*sinA/sinC
而,AB=3+√3= (√3)(1+√3)
sinA = sin60° = √3/2,
sinC = sin(180°-60°-45°) = sin(30°+45°)
= sin30°cos45° + cos30°sin45°
= (1/2)*(√2/2) + (√3/2)*(√2/2)
= √2(1+√3)/4
∴BC = AB*sinA/sinC
= (√3)(1+√3) *(√3/2) /【√2(1+√3)/4】
= 3√2
过点A作AD⊥BC于D
在Rt△ABD中,AD = AB*sin45° = (3+√3)/(√2)
∴S△ABC = (1/2)*BC*AD
= (1/2)*(3√2)*(3+√3)/(√2)
= 3(3+√3)/2
1年前
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