Prove that acosA+bcosB+ccosC=(a+b+c)/2.
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Answered by
10
Sine Rule
a/Sin A = b/SinB = c/sin C
So the above question can be written as
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A
Now using AM- GM inequality (However this can be valid only in the case of acute angled triangels it will not be valid for obtuse angled or right angled triangles )
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= root(a3(CosA CosB Cos C)*(Sin B Sin C)/Sin2A)
withe the equality existing only when the terms r equal
Then on equality aCosA = a (SinBCosB)/SinA = c (SinCCosA)/Sin A
= aSinACosA= aSinBCosB = aCos C SinC
= Sin ACosA = SinBCos B = Sin c Cos C
= A= B = C =60
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A = a/2+b/2+c/2
=> a CosA + B Cos B + Ccos C= (a+b+c)/2.
a/Sin A = b/SinB = c/sin C
So the above question can be written as
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A
Now using AM- GM inequality (However this can be valid only in the case of acute angled triangels it will not be valid for obtuse angled or right angled triangles )
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= root(a3(CosA CosB Cos C)*(Sin B Sin C)/Sin2A)
withe the equality existing only when the terms r equal
Then on equality aCosA = a (SinBCosB)/SinA = c (SinCCosA)/Sin A
= aSinACosA= aSinBCosB = aCos C SinC
= Sin ACosA = SinBCos B = Sin c Cos C
= A= B = C =60
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A = a/2+b/2+c/2
=> a CosA + B Cos B + Ccos C= (a+b+c)/2.
Answered by
5
Sine Rule
a/Sin A = b/SinB = c/sin C
So the above question can be written as
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A
Now using AM- GM inequality (However this can be valid only in the case of acute angled triangels it will not be valid for obtuse angled or right angled triangles )
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= root(a3(CosA CosB Cos C)*(Sin B Sin C)/Sin2A)
withe the equality existing only when the terms r equal
Then on equality aCosA = a (SinBCosB)/SinA = c (SinCCosA)/Sin A
= aSinACosA= aSinBCosB = aCos C SinC
= Sin ACosA = SinBCos B = Sin c Cos C
= A= B = C =60
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A = a/2+b/2+c/2
=> a CosA + B Cos B + Ccos C= (a+b+c)/2.
a/Sin A = b/SinB = c/sin C
So the above question can be written as
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A
Now using AM- GM inequality (However this can be valid only in the case of acute angled triangels it will not be valid for obtuse angled or right angled triangles )
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= root(a3(CosA CosB Cos C)*(Sin B Sin C)/Sin2A)
withe the equality existing only when the terms r equal
Then on equality aCosA = a (SinBCosB)/SinA = c (SinCCosA)/Sin A
= aSinACosA= aSinBCosB = aCos C SinC
= Sin ACosA = SinBCos B = Sin c Cos C
= A= B = C =60
aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A = a/2+b/2+c/2
=> a CosA + B Cos B + Ccos C= (a+b+c)/2.
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