Question

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

• A. ${70}^0$
• B. ${113}^0$
• C. ${123}^0$
• D. ${103}^0$

Answer 1

The correct option is B ${113}^0$

$\angle C={23}^0$
Now, in $\Delta ADC,AC=b$ is the circumdiameter.
Hence, $\frac{AD}{\sin C}=b\Rightarrow \frac{abc}{b^2-c^2}=b\sin C$
$\Rightarrow \frac{\sin A\sin C}{{\sin }^{2}B-{\sin }^{2}C}=\sin C$
$\Rightarrow \frac{\sin A\sin C}{\sin (B+C)\sin (B-C)}=\sin C$
$\Rightarrow \frac{\sin A}{\sin \left(A\right)\sin (B-C)}=1\Rightarrow \sin (B-C)=1\Rightarrow \angle B=\angle C+{90}^0={113}^0$