In triangle ABC,if a=4, b=3, c=2,then 2(a-b c | vedclass

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(D) Given $a=4, b=3, c=2$ in $	riangle ABC$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2 	imes 4 	imes 3} = \frac

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$3$

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(D) Given $a=4, b=3, c=2$ in $	riangle ABC$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2 	imes 4 	imes 3} = \frac{21}{24} = \frac{7}{8}$. $\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{16+4-9}{2 	imes 4 	imes 2} = \frac{11}{16} \implies \sec B = \frac{16}{11}$. Now,substitute these values into the expression: $2(a-b \cos C)(a-c \sec B) = 2(4 - 3 	imes \frac{7}{8})(4 - 2 	imes \frac{16}{11})$ $= 2(4 - \frac{21}{8})(4 - \frac{32}{11})$ $= 2(\frac{32-21}{8})(\frac{44-32}{11})$ $= 2(\frac{11}{8})(\frac{12}{11}) = 2 	imes \frac{12}{8} = 2 	imes \frac{3}{2} = 3$.

In triangle $ABC$,if $a=4, b=3, c=2$,then $2(a-b \cos C)(a-c \sec B) = $

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