find angle b given that c equals 58, a equals 49, and angle a equals 56

Law of Sines:

$\frac{\sin A}{a} = \frac{sin B}{b} = \frac{\sin C}{c}$

Plug in what you know.

$\frac{\sin56}{49} = \frac{\sin C}{58} \\~\\ \frac{58\sin56}{49} = \sin C \\~\\ \arcsin({\frac{58\sin56}{49}}) = C \\~\\ C \approx 78.905^{\circ}$

So since we know that there are 180º in every triangle,

180 ≈ 56 + 78.905 + B

B ≈ 180 - 56 - 78.905

B ≈ 45.095º