I have no problem with right angle triangles, but when it's not, I don't know how to use trigonometry. Because there are $2$ adjacent sides to each angle, I don't know which one to use.
The question is:
A triangles sides are given by coordinates: $A(0,0), B(-2,4), C(4,5)$.
I need to get the angles at each vertex. So I got the length of each side.
$AB$ is: $\sqrt{20}$
BC is: $\sqrt{37}$
AC is: $\sqrt{41}$
I used geogebra for the points, and I got the angles for them, and then I tried sin, cos, tan, none worked, then I tried the cosine function: $c^2=a^2+b^2-2ab\cos(C)$ But I still didn't get the answer.
Could anyone help with solving the question, and explaining in this case what rule to use? Also, how do I know which side is $c$, $a$ and $b$ when the triangle is not right-angled?
By law of cosines $$\cos\measuredangle A=\frac{20+41-37}{2\sqrt20\sqrt{41}}=\frac{6}{\sqrt{205}}.$$ Thus, $$\measuredangle A=\arccos\frac{6}{\sqrt{205}}=65.22...^{\circ}.$$ $\measuredangle B$ and $\measuredangle C$ we can get by the similar way.