Greatest angle

Calculate the greatest triangle angle with sides 197, 208, 299.

Result

α =  95.1 °

Solution:

Δ 197,208,299:Δ 197,208,299  c2=a2+b2−2abcos⁡α  α=arccosa2+b2−c22bc=95.1∘\Delta \ 197, 208, 299: \Delta \ 197, 208, 299 \ \\ \ \\ c^2 = a^2+b^2 - 2ab \cos \alpha \ \\ \ \\ \alpha = arccos \dfrac{ a^2 + b^2 -c^2}{ 2bc} = 95.1 ^\circ

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