Mathematics - Properties of Triangle Question with Solution | TestHub

We know $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$. Substituting these into the expression:$a\cos A+b\cos B+c\cos C=2R\sin A\cos A+2R\sin B\cos B+2R\sin C\cos C=R(\sin 2A+\sin 2B+\sin 2C)$. For any triangle, $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$. So, the expression becomes $4R\sin A\sin B\sin C$. The area of the triangle $\Delta =\frac{abc}{4R}=\frac{(2R\sin A)(2R\sin B)(2R\sin C)}{4R}=2R^2\sin A\sin B\sin C$. From this, $\sin A\sin B\sin C=\frac{\Delta }{2R^2}$. Substituting this back: $a\cos A+b\cos B+c\cos C=4R\left(\frac{\Delta }{2R^2}\right)=\frac{2\Delta }{R}$. Comparing with the given equation $\frac{k\Delta }{R}$, we find $k=2$.