Question

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

• A. ${93}^{\circ }$
• B. ${113}^{\circ }$
• C. ${27}^{\circ }$
• D. ${103}^{\circ }$

Answer 1

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

$\Delta =\frac{1}{2}\times a\times AD$
$\Rightarrow AD=\frac{2\Delta }{a}$
$\Rightarrow \frac{abc}{b^2-c^2}=\frac{2\Delta }{a}\cdots \left(i\right)$
We know,
$\Delta =\frac{abc}{4R}$
putting in equation $\left(i\right)$
$\Rightarrow \frac{abc}{b^2-c^2}=\frac{2\times \frac{abc}{4R}}{a}$
$\Rightarrow \frac{abc}{b^2-c^2}=\frac{bc}{2R}$
$\Rightarrow \frac{a}{b^2-c^2}=\frac{1}{2R}$
Cross multiplying, we get
$\Rightarrow 2Ra=b^2-c^2$
$\Rightarrow 4R^2\sin A=4R^2{\sin }^{2}B-4R^2{\sin }^{2}C$
$\Rightarrow 4R^2\sin A=4R^2\sin (B-C)\sin (B+C)$
$\Rightarrow \sin (B-C)=1$
$\Rightarrow \angle B-\angle C={90}^{\circ }$
$\Rightarrow \angle B={90}^{\circ }+\angle C={90}^{\circ }+{23}^{\circ }={113}^{\circ }$