In a triangle ABC, AD is the altitude from A. Given b>c, $\angle C = {23^o}$,$AD = \frac{{abc}}{{{b^2} - {c^2}}}$. Then $\angle B = \_\_\_{\_^o}$
My approach is as follow
$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$
$AD = \frac{{8{R^3}sinA\sin B\sin C}}{{4{R^2}{{\sin }^2}B - 4{R^2}{{\sin }^2}C}} = \frac{{2RsinA\sin B\sin C}}{{\sin \left( {B - C} \right)\sin \left( {B + C} \right)}} = \frac{{2R\sin B\sin C}}{{\sin \left( {B - C} \right)}}$
How do I approach from here
We can use the extended law of sines and the area of a triangle in terms of sides and circumradius formulae.
$$\bigtriangleup=\frac{1}{2}a\cdot AD \implies AD= \frac{2\triangle}{a} = \frac{{abc}}{{{b^2} - {c^2}}}=\frac{4R \bigtriangleup}{{b^2} - {c^2}}$$
$$\implies 2Ra={b^2} - {c^2}$$ $$\implies 2R(2R\cdot \sin(A))=(2R\cdot \sin(B))^2 - {(2R\cdot \sin(C))}^2$$ $$\implies \sin(A)=\sin^2(B)-\sin^2(C)$$
Can you take it from here?
$$B=C+90°$$
There is also another version of this problem here.
