Question

If in $△ABC,\frac{\sin A}{c}+\frac{\sin C}{a}=\frac{a}{bc}+\frac{c}{ab},$ then the triangle is

• A. right-angled
• B. isosceles
• C. equilateral
• D. none of these.

Answer 1

Answer: (A)

right-angled

$\frac{\sin A}{c}+\frac{\sin C}{a}=\frac{a}{bc}+\frac{c}{ab}$
$\Rightarrow \frac{a\sin A+c\sin C}{ca}=\frac{a^{2}b+b{c}^2}{a{b}^{2}c}$
$\Rightarrow a\sin A+c\sin C=\frac{b(a^2+c^2)}{b^2}$
$\Rightarrow a\sin A+c\sin C=\frac{a^2+c^2}{b}$
$\Rightarrow ab\sin A+bc\sin C=a^2+c^2$
Since $\frac{a}{\sin A}=\frac{c}{\sin C}\Rightarrow \sin A=\frac{a\sin C}{c}$
Substituting $\sin A=\frac{a\sin C}{c}$ above we get
$\Rightarrow ab\left(\frac{a\sin C}{c}\right)+bc\sin C=a^2+c^2$
$\Rightarrow \frac{a^{2}b}{c}\sin C+bc\sin C=a^2+c^2$
$\Rightarrow b(\frac{a^2}{c}+c)\sin C=a^2+c^2$
$\Rightarrow \frac{b}{c}(a^2+c^2)\sin C=a^2+c^2$
$\Rightarrow \frac{b}{c}\sin C=1$
$\Rightarrow \sin C=\frac{c}{b}=\frac{k\sin C}{k\sin B}$
$\therefore \sin B=1\Rightarrow \angle B=\frac{\pi }{2}$