Question

In $△ABC,if\frac{\sin A}{c\sin B}+\frac{\sin B}{c}+\frac{\sin C}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$, then the value of angle A is

• A. ${120}^0$
• B. ${90}^0$
• C. ${60}^0$
• D. ${30}^0$

Answer 1

The correct option is A ${120}^0$

$\text{Solve:-}$

$\text{Given,}$

$\frac{\sin A}{c\sin B}+\frac{\sin B}{C}+\frac{\sin C}{b}=\frac{C}{ab}+\frac{b}{ac}+\frac{a}{bc}$

$\text{according to Sine rule we get,}$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}=2R$

$\Rightarrow \frac{\frac{a}{2R}}{c\times \frac{b}{2R}}+\frac{\sin B}{C}+\frac{\sin C}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$

$\Rightarrow \frac{a}{bc}+\frac{b\sin B+c\sin C}{bc}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$

$=>\frac{b\sin B+c\sin C}{bc}=\frac{c^2+b^2}{abc}$

$\Rightarrow a=\frac{b^2+c^2}{b\sin B+\mathrm{csin}c}$

$\Rightarrow a=\frac{b\cdot \left(2R\sin B\right)+c(2R\sin c}{b\sin B+c\sin c}$

$\Rightarrow a=\frac{2R(b\sin B+c\sin c)}{b\sin B+c\sin c}$

$\Rightarrow a=2R-\left(i\right)$

but,$\frac{a}{\sin A}=2R$

$\Rightarrow a=2R\sin A$ - (ii)

from equation (i) and (ii) we get,

$2R=2R\sin A$

$=>\sin A=1$

$\Rightarrow \angle A={90}^{\circ }$