Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, then $a\cos (B-C)+b\cos (C-A)+c\cos (A-B)$ is equal to?

• A. $\frac{abc}{R^2}$
• B. $\frac{abc}{4R^2}$
• C. $\frac{4abc}{R^2}$
• D. $\frac{abc}{2R^2}$

Answer 1

The correct option is A $\frac{abc}{R^2}$

Formula used:

Sine Rule; $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Projection rule; $c=a\cos B+b\cos A$

We have;

$a\cos (B-C)+b\cos (C-A)+c\cos (A-B)$

$=a(\cos B.\cos C+\sin B.\sin C)+b(\cos C.\cos A+\sin C\sin A)+c(\cos A.\cos B+\sin A.\sin B)$

Putting $\sin A=\frac{a}{2R};\sin B=\frac{b}{2R}\text{and}\sin C=\frac{c}{2R}$

$=a\cos B.\cos C+b\cos C.\cos A+c\cos A.\cos B+\frac{3abc}{4R^2}$

$=\cos C(a\cos B+b\cos A)+c\cos A.\cos B+\frac{3abc}{4R^2}$

By projection rule $(a\cos B+b\cos A)=c$

$=\cos C.c+c\cos A.\cos B+\frac{3abc}{4R^2}$

$=c\left(\cos [\pi -(A+B)]+\cos A.\cos B\right)+\frac{3abc}{4R^2}$

$=c(-\cos (A+B)+\cos A.\cos B)+\frac{3abc}{4R^2}$

$=c.\sin A.\sin B+\frac{3abc}{4R^2}$

Using Sine rule;

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

$=\frac{abc}{4R^2}+\frac{3abc}{4R^2}=\frac{abc}{R^2}$

Hence;

$a\cos (B-C)+b\cos (C-A)+C\cos (A-B)=\frac{abc}{R^2}$