为什么a/sina等于b+c/sinb+sinc
2016-08-25
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∵ 根据正弦定理:a/sinA=b/sinB=c/sinC=2R
∴b=2RsinB,c=2RsinC
∴ (b+c)/(sinB+sinC)=(2RsinB+2RsinC)/(sinB+sinC)
= 2R(sinB+sinC) /(sinB+sinC)
= 2R = a/sinA
∴b=2RsinB,c=2RsinC
∴ (b+c)/(sinB+sinC)=(2RsinB+2RsinC)/(sinB+sinC)
= 2R(sinB+sinC) /(sinB+sinC)
= 2R = a/sinA
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由b/sinB=c/sinC 得 sinC/sinB=c/b 两边加1得 sinC/sinB+1=c/b +1 通分得
(sinC+sinB)/sinB=(b+c)/b 变形得(b+c)/(sinB+sinC) =b/sinB
既b/sinB=c/sinC=(b+c)/(sinB+sinC)
(sinC+sinB)/sinB=(b+c)/b 变形得(b+c)/(sinB+sinC) =b/sinB
既b/sinB=c/sinC=(b+c)/(sinB+sinC)
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a/sinA =b/sinB=c/sinC =k
a= ksinA ,b= ksinB, c=k/sinC
a/sinA = k
(b+c)/(sinB+sinC)
=k(sinB+sinC)/(sinB+sinC)
=k
=a/sinA
a= ksinA ,b= ksinB, c=k/sinC
a/sinA = k
(b+c)/(sinB+sinC)
=k(sinB+sinC)/(sinB+sinC)
=k
=a/sinA
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