Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

What is the value of $a\cos (B-C)+b\cos (C-A)+c\cos (A-B)?$

• 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 $\frac{abc}{R^2}$

Sine formula
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ ....(1)
Consider,
$a\cos (B-C)+b\cos (C-A)+c\cos (A-B)$
$=2R\sin A\cos (B-C)+2R\sin B\cos (C-A)+2R\sin C\cos (A-B)$ (by (1))
$=2R\sin (B+C)\cos (B-C)+2R\sin (C+A)\cos (C-A)+2R\sin (A+B)\cos (A-B)$
$=R\left[2\sin (B+C)\cos \right(B-C)+2\sin (C+A)\cos (C-A)+2\sin (A+B)\cos (A-B\left)\right]$
$=R[\sin 2B+\sin 2C+\sin 2C+\sin 2A+\sin 2A+\sin 2B]$
$=2R[\sin 2A+\sin 2B+\sin 2C]$
$=2R\left(4\sin A\sin B\sin C\right)$ ($\because \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$)
$=8R\left[\frac{a}{2R}\frac{b}{2R}\frac{c}{2R}\right]$ (by (1))
$=\frac{abc}{R^2}$