How to solve for the missing angles of a triangle with sides: a=√3-1, b=2, c=√2

Question

anonymous · Answer

I think you can use the law of Cosine. Just choose any side length to be the side opposite the angle you are trying to find

anonymous · Answer

law of Cosine is c^2=a^2+b^2-2abcosC

elleblythe · Answer

Yes I've tried that but I got a weird answer: -2-2√3+4√2=cosalpha @ArkGoLucky

anonymous · Answer

|dw:1416213154911:dw|
 $$Cos A = (b^2 +c^2 -a^2)/2bc$$
 $$Cos A = (2^2 + (\sqrt{2}^2) -(\sqrt{3} -1)^2)/2*2*\sqrt{2}$$
 $$Cos A = (4 + 2 -(3+1-2\sqrt{3}))/4\sqrt{2}$$
 $$Cos A = (4 -2\sqrt{3})/4\sqrt{2}$$
 $$Cos A = (0.535)/5.65$$
 $$Cos A = 0.094$$
Now Find cos inverse for A

triciaal · Answer

did you have the correct answer if so I have to check my work again if not then mine should be correct. the principle is the same  using the cosine formula