Sine or Cosine Rule?

A solution with sine rule: Using sine rule and area we can easily obtain: $AD=\frac {bc}{2R}=\frac {abc}{b^2-c^2}$ $\implies \frac {b^2-c^2}{a}=2R$ where $R$ is the circumradius of $ABC$ . If $\angle B$ is acute or right, then $\frac {b^2-c^2}{a}=CD-BD=2R\le CD$ which is clearly absurd. Hence $D$ must lie outside of segment $CB$ , and $\frac {b^2-c^2}{a}=CD+BD=2R$ . Let $B'$ be the other point on line $BC$ such that $DB=DB'$ ; thus $CB'=2R$ and $AB=AB'$ . By sine rule again the circumradii of $ABC,AB'C$ are the same, which forces $\angle CAB'=90^{\circ} \implies \angle B=90+\angle C=\boxed {113^{\circ}}$

First, by the Cosine rule, we have that

$c^{2} = a^{2} + b^{2} - 2ab\cos(23^{\circ}) \Longrightarrow b^{2} - c^{2} = a(2b\cos(23^{\circ}) - a)$

$|AD| = \dfrac{abc}{b^{2} - c^{2}} = \dfrac{bc}{2b\cos(23^{\circ}) - a}.$

Now since $AD$ is an altitude we have that $|AD| = b\sin(23^{\circ})$ , so

$b\sin(23^{\circ}) = \dfrac{bc}{2b\cos(23^{\circ}) - a} \Longrightarrow c = 2b\sin(23^{\circ})\cos(23^{\circ}) - a\sin(23^{\circ}).$

Now we also have that $b\sin(23^{\circ}) = |AD| = c\sin(\angle B)$ , so now we have

$c = 2c\sin(\angle B)\cos(23^{\circ}) - a\sin(23^{\circ})$

$\Longrightarrow c(2\sin(\angle B)\cos(23^{\circ}) - 1) = a\sin(23^{\circ})$

$\Longrightarrow \dfrac{a}{c} = \dfrac{2\sin(\angle B)\cos(23^{\circ}) - 1}{\sin(23^{\circ})}.$

But by the Sine rule we know that $\dfrac{a}{c} = \dfrac{\sin(\angle A)}{\sin(23^{\circ})}$ , and so

$2\sin(\angle B)\cos(23^{\circ}) - 1 = \sin(\angle A) = \sin(157^{\circ} - \angle B)$

$\Longrightarrow 2\sin(\angle B)\cos(23^{\circ}) - 1 = \sin(23^{\circ})\cos(\angle B) + \cos(23^{\circ})\sin(\angle B)$

$\Longrightarrow \sin(\angle B)\cos(23^{\circ}) - \cos(\angle B)\sin(23^{\circ}) = 1 \Longrightarrow \sin(\angle B - 23^{\circ}) = 1$

$\Longrightarrow \angle B - 23^{\circ} = 90^{\circ} \Longrightarrow \angle B = \boxed{113^{\circ}}.$

$AD = b\sin(C)$

$\Rightarrow b\sin(C) = \dfrac{abc}{b^2-c^2}$

$\Rightarrow \sin(C) = \dfrac{ac}{b^2-c^2} = \dfrac{\sin(A) \sin(C)}{\sin^{2}(B) - \sin^{2}(C)} = \dfrac{\sin(A)\sin(C)}{\sin(B+C)\sin(B-C)} = \dfrac{\sin(A)\sin(C)}{\sin(A)\sin(B-C)} = \dfrac{\sin(C)}{\sin(B-C)}$

$\Rightarrow \sin(B-C) = 1$

$\Rightarrow B-C = 90^{\circ}$

$B = 113^{\circ}$