Question

In any $\Delta ABC$, $acosA+bcosB+ccosC=$

• A. $\frac{{\Delta }^2}{abc}$
• B. $\frac{4{\Delta }^2}{abc}$
• C. $\frac{8{\Delta }^2}{abc}$
• D. none of these

Answer 1

The correct option is C $\frac{8{\Delta }^2}{abc}$

$acosA+bcosB+ccosC=2R(sinAcosA+sinBcosB+sinCcosC)$
$=R(sin2A+sin2B+sin2C)=R\left(2sin\right(A+B\left)cos\right(A-B)+2sinCcosC)$
$=2RsinC\left(cos\right(A-B)-cos(A+B\left)\right)(\because sin(A+B)=sinC)$
$=2RsinC\left(2sinAsinB\right)=4R\frac{abc}{8R^3}=\frac{abc}{2R^2}$
$=\frac{abc\left(16{\Delta }^2\right)}{2(abc{)}^2}$
$=\frac{8{\Delta }^2}{abc}$