When angle opposite to the unknown sides and other two sides are given then we use law of cosines
Law of cosine
![c^2 = a^2 + b^2 - 2ab cos(c)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/1b05fsskne0y90d8q53yg8259wk5a1ham9.png)
From the given diagram
![AB^2 = CB^2 + AC^2 - 2(CB)(AC) cos( c)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/myxu3qytp9sugq85xrfr3kvuj4hakv1gqk.png)
CB = 108
AC= 55
Angle c= 59
![AB^2 = 108^2 + 55^2 - 2(108)(55) cos( 59 )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4wskfof1m5tdcfyzgmpl60epjm15ufu5t1.png)
![AB^2 = 8570.34767](https://img.qammunity.org/2019/formulas/mathematics/middle-school/97bed9b32cb8109g3kl2azl91bakt33y2m.png)
Take square root on both sides
AB = 92.6 m
To find out angle B we use sine law
![(sin(a))/(a) = (sin b)/(b) = (sin c)/(c)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ttej6qfgukdcpllspebqoje876w403v1d3.png)
![(sin b)/(b) = (sin c)/(c)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/bgcdxyagsc8zvn8zimm4u4tct86dpgerx6.png)
From the figure
![(sin B)/(AC) = (sin C)/(AB)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ub9pirfnd7qhstux3lks68rlrkp2euwnj3.png)
![(sin B)/(55) = (sin 59)/(92.6)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/rz19yt7f4o0aatjuknqpcthv8z1pyp2bln.png)
sin(B) = 0.50916647
B =
(0.50916647)
Angle B= 30.61 degrees