Question

In a triangle $A B C, A D$ is the altitude from $A$. If $b>c$, $\angle C=23^{\circ}$ and $A D=\frac{a b c}{b^{2}-c^{2}}$ then $\angle B=$

• A. $87^{\circ}$
• B. $113^{\circ}$
• C. $118^{\circ}$
• D. none of these

Answer 1

Answer: (B)

$\Delta=\frac{1}{2} a \cdot A D$
$\Rightarrow A D=\frac{2 \Delta}{a}$
Also, $A D=\frac{a b c}{b^{2}-c^{2}}$ (Given)
$\Rightarrow \frac{2 \Delta}{a}=\frac{4 R \cdot \Delta}{b^{2}-c^{2}}$
$\Rightarrow 2 R a=b^{2}-c^{2}$
$\Rightarrow \sin A=\sin ^{2} B-\sin ^{2} C$
$=\sin (B+C) \cdot \sin (B-C)$
$=\sin A \cdot \sin (B-C)$
$\Rightarrow \sin (B-C)=1$
$\Rightarrow \angle B-\angle C=\frac{\pi}{2}$
$\Rightarrow \angle B=\frac{\pi}{2}+\angle C$
$\Rightarrow \angle B=90^{\circ}+23^{\circ}=113^{\circ}$