Let's say given an angle A = 46 °, side a = 2.29 and b = 2.71
I figured that the angle B = 58.4 by saying:
$$B = \sin^{-1} \left(\frac{ 2.71 \sin{46^{\circ}}}{2.29}\right)=58.4^{\circ}$$
But I think that angle C is incorrect:
$$C = \sin^{-1} \left(\frac{2.29 \sin{58.4^{\circ}}}{2.71}\right)=46.03^{\circ}$$
Someone who can help me? what do I do wrong and how should it be done?
Best Answer
While the other answers have covered the basic use of the Law of Sines, they've all missed a critical point: there are two triangles that fit your given information. Here is the triangle you've found:
But, when you're solve $\sin B=\frac{b\sin A}{a}$, there are two solutions that could be angles in a triangle, $B_1=\arcsin\left(\frac{b\sin A}{a}\right)$ (the one you found) and $B_2=180°-\arcsin\left(\frac{b\sin A}{a}\right)$. Using this second possible measure for $B$ won't always yield a triangle, but in this case it does:
Here are both triangles pictured together:
This issue with the Law of Sines, sometimes called the "ambiguous case," is almost certainly what the picture in your question here is about.
See also my answer here on general techniques for triangle-solving.