Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

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 B ${90}^0$

From Sin law $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{\sin A}{c\sin B}+\frac{\sin B}{c}+\frac{\sin C}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$
$\Rightarrow \frac{a\sin A}{cb\sin A}+\frac{b\sin A}{ac}+\frac{c\sin A}{ab}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$
Only when $\sin A=1\Rightarrow A=90^0$