Using Law of sines SSA ( side side angle)

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\~\\ \frac{\sin 56}{17} = \frac{\sin C}{14} \\~\\ \frac{14\sin 56}{17} = \sin C \\~\\ \arcsin{(\frac{14\sin 56}{17})} = C \\~\\ C \approx 43.058^{\circ}$

Since there are 180 degrees in every triangle,

180 = A + B + C

A = 180 - B - C

A ≈ 180 - 56 - 43.058

A ≈ 80.942º

All that's left to find is side a.

$\frac{\sin 56}{17} \approx \frac{\sin 80.942}{a} \\~\\ (a)\frac{\sin 56}{17} \approx \sin 80.942 \\~\\ a \approx \frac{17\sin80.942}{sin56} \\~\\ a \approx 20.250$

Recap:

A ≈ 80.942º          a ≈ 20.25

B = 56º                   b = 17

C ≈ 43.058º            c = 14

*edit* I put c = 4 instead of c = 14

triangle was A,B,C angle B 56 and side length little c is 14 little b is 17 solve triangle using Law of sines (SSA case)

Solving for angle C we have

sin C / 14  = sin 56 / 17

sin C  = [14 * sin 56] / 17  

arcsin [ (14 * sin 56) / 17 ]  ≈ 43°

And angle A   =   180 - 56 - 43  ≈   81°

And  side a  =

a / sin 81  =  17/ sin 56   →  a = 17sin 81/sin 56 ≈ 20.25

To see if we have a second triangle, C might also  be  180 - 43  = 137°

However......when we add this to the known angle of 56°, we get more than 180°....so....only one triangle is possible

a = 20.25        b = 17       c  =  14

A = 81            B = 56       C = 43