Question

求大神帮忙做一下数学题

Answer 1

已知向量m、n互相垂直，则：
sinB(5sinB-6sinC)+(5sinA+5sinC)(sinC-sinA)=0
==> 5sin²B-6sinBsinC+5sin²C-5sin²A=0
==> 5b²-6bc+5c²-5a²=0
==> 5(b²+c²-a²)=6bc
==> (b²+c²-a²)/2bc=3/5
==> cosA=3/5
所以，sinA=4/5

由正弦定理有：a/sinA=b/sinB=c/sinC
==> 2√2/(4/5)=b/sinB=c/sinC
==> b=(5√2/2)sinB，c=(2√5/2)sinC
则，S△ABC=(1/2)bcsinA=(1/2)×(2√5/2)²*sinB*sinC×(4/5)=5sinBsinC
=(-5/2)[cos(B+C)-cos(B-C)]
=(5/2)[cos(B-C)-cos(B+C)]
=(5/2)[cos(B-C)+cosA]
=(5/2)[cos(B-C)+(3/5)]
所以，当B=C时，上式有最大值=(5/2)[1+(3/5)]=4

Answer 2

利用两个向量的数量级为0，然后化简即可

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