Respuesta :

For the given triangle, we must apply the following trigonometric relation:

[tex]\begin{gathered} k^2=j^2+l^2-2jl\cos K \\ \cos K=\frac{k^2-j^2-l^2}{-2jl} \\ \cos K=\frac{12^2-13^2-10^2}{-2\cdot13\cdot10} \\ \cos K=0.48 \end{gathered}[/tex]

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