1. ## trigo proof

in triangle ABC if (acosA+bcosB+ccosC)/(asinB+bsinC+csinA)=(a+b+c)/9R where R is circumradius then prove that the triangle is equilateral

2. You need to show some working, or share your thoughts before we can help you. How do you think you should start this question?

3. i thick we should use the independant values of cos A cos B etc

4. Originally Posted by prasum
in triangle ABC if (acosA+bcosB+ccosC)/(asinB+bsinC+csinA)=(a+b+c)/9R where R is circumradius then prove that the triangle is equilateral
One way to go about it is to obtain expressions for $R$ and $aCosA+bCosB+cCosC$

$aCosA+bCosB+cCosC=2aSinBSinC$

$R=\displaystyle\frac{abc}{2r(a+b+c)}$ where $r$ is the incircle radius.

If you draw perpendicular lines from the incentre to all 3 sides, then

$\displaystyle\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{ 2}ar=\frac{1}{2}abSinC=\frac{1}{2}cbSinA=\frac{1}{ 2}acSinB=area$

$\Rightarrow\ r(a+b+c)=abSinC$

$\displaystyle\Rightarrow\ R=\frac{abc}{2abSinC}=\frac{c}{2SinC}$

Therefore, if

$\displaystyle\frac{aCosA+bCosB+cCosC}{asinB+bSinC+ cSinA}=\frac{a+b+c}{9R}$

then

$\displaystyle\ (9)\;\frac{2aSinBSinC}{aSinB+bSinC+cSinA}=\frac{a+ b+c}{R}=\frac{(a+b+c)2SinC}{c}$

$\displaystyle\Rightarrow\frac{9aSinB}{aSinB+bSinC+ cSinA}=\frac{a+b+c}{c}$

$\Rightarrow\ 9acSinB=(a+b+c)(aSinB+bSinC+cSinA)$

and

$\displaystyle\ 9acSinB=(18)\;\frac{1}{2}acSinB$

is 18 times the triangle area.

Therefore

$(a+b+c)(aSinB+bSinC+cSinA)=18(area)\;\;\Rightarrow \ a=b=c$

and so the triangle is equilateral.

5. Originally Posted by Archie Meade
One way to go about it is to obtain expressions for $R$ and $aCosA+bCosB+cCosC$

$aCosA+bCosB+cCosC=2aSinBSinC$

$R=\displaystyle\frac{abc}{2r(a+b+c)}$ where $r$ is the incircle radius.

If you draw perpendicular lines from the incentre to all 3 sides, then

$\displaystyle\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{ 2}ar=\frac{1}{2}abSinC=\frac{1}{2}cbSinA=\frac{1}{ 2}acSinB=area$

$\Rightarrow\ r(a+b+c)=abSinC$

$\displaystyle\Rightarrow\ R=\frac{abc}{2abSinC}=\frac{c}{2SinC}$

Therefore, if

$\displaystyle\frac{aCosA+bCosB+cCosC}{asinB+bSinC+ cSinA}=\frac{a+b+c}{9R}$

then

$\displaystyle\ (9)\;\frac{2aSinBSinC}{aSinB+bSinC+cSinA}=\frac{a+ b+c}{R}=\frac{(a+b+c)2SinC}{c}$

$\displaystyle\Rightarrow\frac{9aSinB}{aSinB+bSinC+ cSinA}=\frac{a+b+c}{c}$

$\Rightarrow\ 9acSinB=(a+b+c)(aSinB+bSinC+cSinA)$

and

$\displaystyle\ 9acSinB=(18)\;\frac{1}{2}acSinB$

is 18 times the triangle area.

Therefore

$(a+b+c)(aSinB+bSinC+cSinA)=18(area)\;\;\Rightarrow \ a=b=c$

and so the triangle is equilateral.
thanks

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# PROVE THAT acosA bcosB ccosC=2asinBsinC

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