Trigo bash-3

$\begin{aligned} \sin B & = \frac {AD}c \\ & = \frac {ab}{b^2-c^2} & \small \color{#3D99F6} \text{By cosine rule: }c^2 = a^2 + b^2 - 2ab\cos C \\ & = \frac b{2b\cos C - a} & \small \color{#3D99F6} \text{By sine rule: }\frac {\sin A}a = \frac {\sin B}b = \frac {\sin C}c \\ & = \frac {\frac {c\sin B}{\sin C}}{2 \frac {c\sin B \cos C}{\sin C} - \frac {c\sin A}{\sin C}} \\ & = \frac {\sin B}{2\sin B \cos C- \sin A} \end{aligned}$

$\begin{aligned} \implies 2 \sin B \cos C - \color{#3D99F6} \sin A & = 1 & \small \color{#3D99F6} \text{Note that }\sin A = \sin (180^\circ - B-C) = \sin (B+C) \\ 2 \sin B \cos C - \color{#3D99F6} \sin (B+C) & = 1 \\ 2 \sin B \cos C - \sin B\cos C - \cos B \sin C & = 1 \\ \sin B \cos C - \cos B \sin C & = 1 \\ \sin (B-C) & = 1 \\ \implies B - C & = 90^\circ \\ B & = 90^\circ + 23^\circ \\ & = \boxed{113}^\circ \end{aligned}$