Question

A triangle has sides of 2,3 and 4 using the law of cosines , a^2+b^2-2abcosC=c^2 what is the value of 2abcosC

Answer 1

`\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\qquad \begin{cases} a=2\\ b=3\\ c=4 \end{cases} \\\\\\ 4^2=2^2+3^2-2abcos(C)\implies 2abcos(C)=2^2+3^2-4^2`