Question

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

• A. ${113}^o$
• B. ${123}^o$
• C. ${147}^o$
• D. ${157}^o$

Answer 1

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

In $\Delta ADC$
$AD=bsinC$
$\Rightarrow \frac{abc}{b^2-c^2}=bsinC$
$\Rightarrow \frac{sinC}{c}=\frac{a}{b^2-c^2}$
$\frac{sinA}{a}=\frac{a}{b^2-c^2}$
$\Rightarrow sinA=\frac{a^2}{b^2-c^2}[\because \frac{sinC}{c}=\frac{sinA}{a}]$
$\Rightarrow \sin A=\frac{si{n}^{2}A}{si{n}^{2}B-si{n}^{2}C}$
$\Rightarrow \sin A=\frac{si{n}^{2}A}{sin(B+C)sin(B-C)}$
$\Rightarrow sin(B-C)=1[\because sin(B+C)=sinA]$
$\Rightarrow B-C={90}^o$
$\Rightarrow B={90}^o+C={90}^o+{23}^o={113}^o$