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In triangle JKL j=10cm k=12cm anf l=13cm find cos K

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


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

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