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

1 answer:

7 0

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

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