If \((\mathrm{a}+\mathrm{b}) \cos \mathrm{C}+(\mathrm{b}+\mathrm{c}) \cos \mathrm{A}+(\mathrm{c}+\mathrm{a}) \cos \mathrm{B}=72\) and if \(\mathrm{a}=18, \mathrm{~b}=24\), then area of the triangle ABC is

\(\begin{aligned} &(a+b) \cos C+(b+c) \cos A+(c+a) \cos B=72 \\ & \therefore \quad a \cos C+b \cos C+b \cos A+c \cos A \\ &+c \cos B+a \cos B=72 \\ & \Rightarrow a \cos C+c \cos A+b \cos A+a \cos B \\ &+b \cos C+c \cos B=72\end{aligned}\)
\(\cdots\left[\begin{array}{l}\text { By projection } \\ a=b \cos C+c \cos B \\ b=c \cos A+a \cos C \\ c=a \cos B+b \cos A\end{array}\right]\)
\(\begin{aligned} & \Rightarrow \mathrm{b}+\mathrm{c}+\mathrm{a}=72 \\ & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=72 \\ & \begin{aligned} & \Rightarrow 18+24+\mathrm{c}=72 \quad[\mathrm{a}=18, \mathrm{~b}=24 \ldots . \text { Given }] \\ & \Rightarrow \mathrm{c}=30\end{aligned} \\ & \begin{aligned} \text { Area of } \triangle \mathrm{ABC} & =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\ & =\sqrt{36(36-18)(36-24)(36-30)} \\ & =\sqrt{36 \times 18 \times 12 \times 6} \\ & =216 \text { sq. units }\end{aligned}\end{aligned}\)