Question

In $\Delta ABC,if\frac{sinA}{csinB}+\frac{sinB}{c}+\frac{sinC}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc},$ then the value of angle A, is (all symbols used have their usual meaning in a triangle):

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

Answer 1

The correct option is B ${90}^0$

We have,

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

We know that

Using $SINErule$ and we get,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

$\sin A=ak......\left(1\right)$

$\sin B=bk.......\left(2\right)$

$\sin C=ck.......\left(3\right)$

Now,

$\frac{ak}{ckb}+\frac{kb}{c}+\frac{kc}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$

$\frac{a}{bc}+\frac{kb}{c}+\frac{kc}{b}=\frac{c}{ab}+\frac{b}{ac}+\frac{a}{bc}$

$\frac{kb}{c}+\frac{kc}{b}=\frac{c}{ab}+\frac{b}{ac}$

$k(\frac{b}{c}+\frac{c}{b})=\frac{1}{a}(\frac{c}{b}+\frac{b}{c})$

$k=\frac{1}{a}$

Put the value of k in equation (1) and we get,

$\sin A=a\frac{1}{a}$

$\sin A=1$

$\sin A=\sin {90}^o$

$A={90}^o$

Hence, this is the answer.