Question

In $△ABC,AD$ is an altitude from $A$ on side $BC$. If $b>c,\angle C={23}^{\circ }$ and $AD=\frac{abc}{b^2-c^2}$, then $\angle B$ is

• A. ${113}^{\circ }$
• B. ${110}^{\circ }$
• C. ${117}^{\circ }$
• D. ${112}^{\circ }$

Answer 1

The correct option is A ${113}^{\circ }$

In $△ABC$, given that $AD$ is an altitude from $A$ on $BC$,

$b>c,\angle C={23}^{\circ }$ and $AD=\frac{abc}{b^2-c^2}$


$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{a}{2c}-\frac{b^2-c^2}{2ac}$
$=\frac{a}{2c}-\frac{b^2-c^2}{abc}\cdot \frac{b}{2}=\frac{a}{2c}-\frac{b}{2AD}=\frac{\sin A}{2\sin C}-\frac{1}{2\sin C}$

In $△ACD$
$\{\frac{AD}{b}=\sin C\Rightarrow \frac{b}{AD}=\frac{1}{\sin C}\}$
$\Rightarrow \cos B=\frac{\sin A-1}{2\sin C}$

$\Rightarrow 2\cos B\sin C=\sin A-1$
$\Rightarrow \sin (B+C)-\sin (B-C)=\sin A-1$

$\Rightarrow \sin A-\sin (B-C)=\sin A-1$
$\Rightarrow \sin (B-C)=1$
$\Rightarrow B-C=\frac{\pi }{2}$
$\Rightarrow B=C+\frac{\pi }{2}={23}^{\circ }+{90}^{\circ }={113}^{\circ }$