Question

In a triangle $ABC,AD$ is the altitude from$A$.Given $b>c,\angle C={23}^0$ and $AD=\frac{abc}{b^2-c^2},$ then $\angle B$ is equal to

• A. ${53}^0$
• B. ${113}^0$
• C. ${87}^0$
• D. none of these

Answer 1

Answer: (B)

${113}^0$

$\because AD=\frac{abc}{b^2-c^2}$
$=\frac{k\sin A.b.k\sin C}{k^2({\sin }^{2}A-{\sin }^{2}C)}$
$=\frac{b\sin A\sin C}{\sin (B+C)\sin (B-C)}$
$=\frac{b\sin A\sin C}{\sin (\pi -A)\sin (B-C)}$
$=b\sin C$ from $△ADC$
$\therefore \sin (B-C)=1$
$\therefore B-C={90}^0$
$\therefore B={90}^0+C={90}^0+{23}^0={113}^0$