Question

In $△ABC$, if $a\cos A+b\cos B+c\cos C=\frac{2\Delta }{k}$ then $k=$

• A. $r$
• B. $R$
• C. $s$
• D. $R^2$

Answer 1

Answer: (B)

$R$

To Prove $a\cos A+b\cos B+c\cos C=\frac{2△}{k}$

L.H.S $=a\cos A+b\cos B+c\sin C$

In any triangle ABC

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ (sine rule)

$\Rightarrow a=2R\sin A,$ ----

$\Rightarrow L.H.S=2R\sin A\cos A+2R\sin B\cos B+2R\sin C\cos C$

$=R(\sin 2A+\sin 2B+\sin 2C)$

In a $△ABC,$

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ $\dots [\because$ conditional identity $]$

$\Rightarrow L.H.S=R\left(4\sin A\sin B\sin C\right)$

$=4R\sin A\sin B\sin C$

From sine rule $\sin B=\frac{b}{2R},\sin A=\frac{a}{2R}$

$\Rightarrow L.H.S=4R\left(\frac{a}{2R}\right)\left(\frac{b}{2R}\right)\sin c$

$=\frac{1}{R}ab\sin c$

$=\frac{2}{R}\left(\frac{1}{2}ab\sin c\right)$

area $△=\frac{1}{2}ab\sin c$

$\Rightarrow L.H.S=\frac{2}{R}△=\frac{2△}{R}$

$R.H.S=\frac{2△}{k}$

$\Rightarrow k=R$