Question

In a triangle $ABC,AD$ is the altitude from $A$, Given $b>c,\angle C={23}^o$ and $AD=\frac{abc}{b^2-c^2},$

then $\angle B=...........$

Answer 1

Ans. ${90}^o+{23}^o={113}^o$, BY sine rule

$AD=c\sin B=\frac{\sin A\sin B.c}{{\sin }^{2}B-{\sin }^{2}C}$

$\therefore \sin (B+C)\sin (B-C)=\sin A$

or $\therefore \sin (B-C)=1=\sin {90}^o$

$\therefore B={90}^o+C$