Respuesta :
thats better thank you
In this case we must use the cosines theorem
[tex]\begin{gathered} a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos A \\ FIND\text{ A} \\ A=\cos ^{-1}(\frac{b^2+c^2-a^2}{2\cdot b\cdot c}) \end{gathered}[/tex]now use this formula to find every angle
[tex]\begin{gathered} a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos A \\ FIND\text{ B} \\ B=\cos ^{-1}(\frac{130^2+95^2-110^2}{2\cdot130\cdot95}) \\ B\approx55.96 \end{gathered}[/tex]DO THE SAME FOR THE OTHER 3
[tex]\begin{gathered} a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos A \\ FIND\text{ N} \\ N=\cos ^{-1}(\frac{130^2+110^2-95^2}{2\cdot130\cdot110}) \\ N\approx45.70 \end{gathered}[/tex]LASTLY P
[tex]\begin{gathered} a^2=b^2+c^2-2\cdot b\cdot c\cdot\cos A \\ FIND\text{ P} \\ P=\cos ^{-1}(\frac{110^2+95^2-130^2}{2\cdot110\cdot95}) \\ P\approx78.34 \end{gathered}[/tex]The greatest angle will be P
