Question

In a triangle $ABC$, AD is the altitude from $A$. Given $b>c$, $\angle C={23}^o$ and $AD=\frac{abc}{b^2-c^2}$; then $\angle B$ equal to ?

• A. ${67}^o$
• B. ${113}^o$
• C. ${44}^o$
• D. $Noneofthese$

Answer 1

The correct option is B ${113}^o$

$AD=\frac{abc}{b^2-c^2}$

$\Rightarrow AD=\frac{k\sin A\cdot b\cdot k\sin C}{k^2({\sin }^{2}B-{\sin }^{2}C)}\left[\text{Using sine law}\right]$

$\Rightarrow AD=\frac{b\cdot \sin A\cdot \sin C}{{\sin }^{2}B-{\sin }^{2}C}$

$\Rightarrow AD=\frac{b\cdot \sin A\cdot \sin C}{\sin (B+C)\cdot \sin (B-C)}$

$\Rightarrow AD=\frac{b\cdot \sin C}{\sin (B-C)}[\because A+B+C=\pi ]$

But,

$AD=b\sin C$

Therefore,

$\sin (B-C)=1$

$\Rightarrow B-C=90$

$\Rightarrow B=90+23=113[\because \angle C=23]$