Problem
Solution
For this case we can use the Cosine law and we have:
[tex]\cos (c)=\frac{a^2+b^2-c^2}{2ab}=\frac{35^2+56^2-33^2}{2\cdot35\cdot56}=0.8346938776[/tex][tex]\cos (b)=\frac{a^2+c^2-b^2}{2ac}=\frac{35^2+33^2-56^2}{2\cdot35\cdot33}=-0.356[/tex][tex]\cos (a)=\frac{c^2+b^2-a^2}{2cb}=\frac{33^2+56^2-35^2}{2\cdot33\cdot56}=0.8116883117[/tex]
And then we can find the angles with the arcos like this:
[tex]<\gamma=ar\cos (0.8346938776)=33.42[/tex][tex]<\beta=ar\cos (-0.356)=110.85[/tex][tex]<\alpha=ar\cos (0.8116883117)=35.74[/tex]
Can you see the new values for gamma and alfa
