Question

如图,在△ABC中,∠B=45°,AC=5,BC=2根号2,求sinA和AB

Answer 1

据正弦定理：因为BC/sinA=AC/sinB
所以sinA=BC×sinB/AC
所以sinA=2√2×√2/2/5=2/5
据余弦定理：AC²=AB²+BC²-2AB·BC·cosB
代入AC,BC,cosB
所以AB=2+√21

Answer 2

由正弦定理可知,2根号2/sinA=5/sinB,所以可得sinA=2/5,由余弦定理可得,AB的平方=AC的平方-BC的平方+2AB乘BC乘cosB,代入数据,解得AB=(4+根号67)/2

Answer 3

 

AD=x

DC=2√2-x

AD²+DC²=AC²

x²+(2√2-x)²=5²

x=√2+√42/2

AB=AD*√2=2+√21

S⊿ABC=1/2*AD*BC=1/2AB*AC*sinA

sinA=BC*AD/(AB*AC)=2/5

Answer 4

2√2/sinA=5/sin45°,sinA=2√2*√2/2/5=4/5,,cosA=3/5,cosB=sinB=√2/2
sinC=sin[180°-(A+B)]=sin(A+B)=sinAcosB+sinBcosA=4/5*√2/2+√2/2*3/5=7√2/10
AB/(7√2/10)=5/(√2/2), AB=7

Answer 5

BC/sinA=AC/sin45°
sinA=BC*sin45°/AC=2*(√2/2)/5=√2/5
AB/sinC=AC/sinB
AB=AC*sinC/sinB
sinC=sin(180°-45°-A)
=sin(135°-A)
=sin135°cosA-cos135°sinA
由于BC<AC ∠A<∠B
cosA=√(1-(√2/5)^2)=√23/5
sin135°=√2/2 cos135°=-√2/2
sinC=√2/2(√23/5+√2/5)
AB=5*√2/2(√23/5+√2/5)/(√2/2)
=√23+√2