Again the same

Group terms and write them as: $\small{(\color{#D61F06}{a\cos B+b\cos A})+(\color{#3D99F6}{a\cos C+c\cos A})+(\color{#20A900}{b\cos C+c\cos B})}$

$=\color{#D61F06}{c}+\color{#3D99F6}{b}+\color{#20A900}{a}$

By Vieta's formula this is

$\Large =\boxed{24}$

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Proof:- $a\cos B+b\cos A=c$

Recall circumradius $R$ . Write $a=2R\sin A,b=2R\sin B$ and then use $\sin A\cos B+\cos A\sin B=\sin (A+B)=\sin C$ so that we are left with $2R\sin C$ which is nothing but $c$ . In a similar way we can prove other two also so that :

$(\color{#D61F06}{a\cos B+b\cos A})=c\\(\color{#3D99F6}{a\cos C+c\cos A})=b\\(\color{#20A900}{b\cos C+c\cos B})=a$

Much popularly known as Projection formula .

If you didn't know the projection formula (like me):

From Vieta's formula , we have:

$\color{#3D99F6}{a+b+c=24}\\ \color{#D61F06}{ab+ac+bc=180}\\ \color{#EC7300}{abc=420}$

From cosine rule , we also know that

$a^2=b^2+c^2-2bc \cos A\\ \implies \cos A = \dfrac{b^2+c^2-a^2}{2bc}$

Similarly, we can find that

$\cos B = \dfrac{a^2+c^2-b^2}{2ac}\\ \cos C = \dfrac{a^2+b^2-c^2}{2ab}$

Therefore,

$a(\cos B+\cos C)+b(\cos C+\cos A)+c(\cos A+\cos B)\\ =\color{magenta}{\cancel{\color{#333333}{a}}}\left(\dfrac{a^2+c^2-b^2}{2\color{magenta}{\cancel{\color{#333333}{a}}}c}+ \dfrac{a^2+b^2-c^2}{2\color{magenta}{\cancel{\color{#333333}{a}}}b}\right)+\color{magenta}{\cancel{\color{#333333}{b}}}\left( \dfrac{a^2+b^2-c^2}{2a\color{magenta}{\cancel{\color{#333333}{b}}}}+ \dfrac{b^2+c^2-a^2}{2\color{magenta}{\cancel{\color{#333333}{b}}}c}\right)+\color{magenta}{\cancel{\color{#333333}{c}}}\left( \dfrac{b^2+c^2-a^2}{2b\color{magenta}{\cancel{\color{#333333}{c}}}}+\dfrac{a^2+c^2-b^2}{2a\color{magenta}{\cancel{\color{#333333}{c}}}}\right)\\ =\color{#20A900}{\dfrac{a^2+c^2-b^2}{2c}}+ \color{#69047E}{\dfrac{a^2+b^2-c^2}{2b}}+\color{#624F41}{\dfrac{a^2+b^2-c^2}{2a}}+ \color{#20A900}{\dfrac{b^2+c^2-a^2}{2c}}+\color{#69047E}{\dfrac{b^2+c^2-a^2}{2b}}+\color{#624F41}{\dfrac{a^2+c^2-b^2}{2a}}\\ =\color{#20A900}{\dfrac{2c^2}{2c}}+ \color{#69047E}{\dfrac{2b^2}{2b}}+\color{#624F41}{\dfrac{2a^2}{2a}}\\ =\color{#3D99F6}{a+b+c}\\ =\boxed{\color{#3D99F6}{24}}$

This is called the projection rule. Its proof is very simple

Call the triangle to ABC And opposite sides be a,b,c .

drop a perpendicular from A Onto BC Call foot of perpendicular to be D.

Now BD+DC = BC (BC = ccos(B) And DC = bcos(C)

Hence bcosc+ccos(B) = a

So the expression asked is nothing but a+b+c which can be found out using vieta's