In a triangle ABC, AD is the altitude from A (Fig. 12.11). Given b > c, $\angle$ C = 23° and $AD=\frac{abc}{b^2-c^2}$then $\angle$ B. is equal to 

                                          

We have AD = b sin C

$\begin{array}{l}\Rightarrow \frac{abc}{b^2-c^2}=b\sin C\\ \Rightarrow \frac{\sin C}{c}=\frac{a}{b^2-c^2}\\ \Rightarrow \frac{\sin A}{a}=\frac{a}{b^2-c^2}\\ \Rightarrow \sin A=\frac{a^2}{b^2-c^2}\left[\because \frac{\sin C}{c}=\frac{\sin A}{a}\right]\end{array}$

$\begin{array}{l}=\frac{{\sin }^2 A}{{\sin }^2 B-{\sin }^2 C}\\ =\frac{{\sin }^2 A}{\sin (B+C)\sin (B-C)}\end{array}$

$\begin{array}{l}\sin (B-C)=1\\ B-C={90}^{\circ }[\because \sin (B+C)=\sin A]\end{array}$

$\Rightarrow B=90°+C=90°+23°=113°$.