Question

在三角形ABC中，B=45,a=4倍根号3,b=4倍2，求A,C,c

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Answer 1

B=45°
a=4√3
b=4√2
根据正弦定理：a/sinA=b/snB
sinA = asinB/a = 4√3 * sin45°/(4√2) = √3/2
A=60°或120°
C=180°-A-B=75°，或15°
做CD⊥AB于D
当A=60°，C=75°时，c = BD + AD = acosB+ bcosA = 4√3cos45 ° + 4√2*cos60° = 2(√6+√2)
当A=120°，C=15°时，c = BD - AD = acosB - bcosA = 4√3cos45 ° - 4√2*cos60° = 2(√6-√2)

Answer 2

根据正弦定理
a/sinA=b/sinB=c/sinC
4倍根号2/sin45°=4倍根号3/sinA
sinA=根号3/2
所以A=60°或120°
当A=60°
C=180°-A-B=75°
sin75°
=sin（45°+30°）
=sin45°×cos30°+cos45°×sin30°
=√2/2×√3/2+√2/2×1/2
=√6/4+√2/4
=（√2+√6）/4
c/sinC=a/sinA
c=a/sinA*sinC=4倍根号3/(根号3/2)*[（√2+√6）/4]=2*（√2+√6）
当A=120°
C=180°-A-B=15°
sin15=sin(45-30)
=sin45cos30-sin30cos45
=√2/2×√3/2-1/2×√2/2
=√6/4-√2/4
=（√6-√2）/4
c/sinC=a/sinA
c=a/sinA*sinC=4倍根号3/(根号3/2)*[（√6+√2）/4]=2*（√6-√2）