Question

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

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

Answer 1

Answer: (B)

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