Question

In $△ABC$, $a\cos A+b\cos B+c\cos C=\frac{8{\Delta }^2}{abc}$

• A. True
• B. False

Answer 1

The correct option is A True

By using sine rule, we can obtain values of a, b and c and then by substituting these values in L.H.S. we can prove this.

$⟹\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=k$ (let)[by sine rule]

Then, $aksinA,b=ksinB$ and $c=ksinC$

Now, $acosA+bcosB+ccosC$

$ksinAcosA+ksinBcosB+ksinCsinC$

$\frac{k}{2\left[sin2AsinBsinC\right]}=2k\left[sinAsinBsinC\right]$

$⟹2asinBsinC=2a.\left(\frac{2\Delta }{ac}\right)\left(\frac{2\Delta }{ab}\right)$

$\therefore \Delta =\frac{1}{2}absinC=\frac{1}{2}acsinB$

$⟹sinB=\frac{2\Delta }{bc},sinC=\frac{2\Delta }{ab}$

$\Rightarrow =\frac{8{\Delta }^2}{abc}=R.H.S$