Question

In a triangle $ABC,AD$ is altitude from $A$. If $b>c,C={23}^0,AD=\frac{abc}{b^2-c^2}$, then $B=$

• A. ${107}^0$
• B. ${97}^0$
• C. ${117}^0$
• D. ${113}^0$

Answer 1

The correct option is D ${113}^0$

$c\sin B=AD=\frac{abc}{b^2-c^2}$
$\Rightarrow \sin B(b^2-c^2)=ab$
$\Rightarrow \sin B({\sin }^{2}B-{\sin }^{2}C)=ab$ using sine rule
$\Rightarrow \sin B\left[\sin (B+C)\sin (B-C)\right]=\sin A\sin B$ by sine rule
$\Rightarrow \sin (B+C)\sin (B-C)=\sin A$
$\Rightarrow \sin (B+C)\sin (B-C)=\sin (\pi -(B+C))$
$\Rightarrow \sin (B+C)\sin (B-C)=\sin (B+C)$
$\Rightarrow \sin (B-C)=1$
$\Rightarrow B-C={90}^0$
Given $C={23}^0$
$\Rightarrow B={90}^0+C={90}^0+{23}^0={113}^0$