If (a+b) cos C + (b+c) cos A + (c+a) cos B = | vedclass

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(B) Given the equation: $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$ Expanding the terms: $a \cos C + b \cos C + b \cos A + c \cos A + c \cos B +

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$216 	ext{ sq. units}$

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(B) Given the equation: $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$ Expanding the terms: $a \cos C + b \cos C + b \cos A + c \cos A + c \cos B + a \cos B = 72$ Rearranging the terms: $(a \cos C + c \cos A) + (b \cos A + a \cos B) + (b \cos C + c \cos B) = 72$ Using the projection formula: $b = c \cos A + a \cos C$,$c = a \cos B + b \cos A$,and $a = b \cos C + c \cos B$ Substituting these into the equation: $b + c + a = 72$ Given $a = 18$ and $b = 24$: $18 + 24 + c = 72$ $\Rightarrow 42 + c = 72$ $\Rightarrow c = 30$ The semi-perimeter $s = \frac{a+b+c}{2} = \frac{72}{2} = 36$ Using Heron's formula: $	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ $	ext{Area} = \sqrt{36(36-18)(36-24)(36-30)} = \sqrt{36 	imes 18 	imes 12 	imes 6}$ $	ext{Area} = \sqrt{36 	imes 1296} = 6 	imes 36 = 216 	ext{ sq. units}$

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

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