Discussion Forum

 In $\triangle$ ABC, with usual  notations  , if   $\cos A=\frac{\sin B}{\sin C'}$ then the triangle  is ........

Answer:

Explanation:

 Key Idea  Use sine rule,

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

 we have,

   $\cos A=\frac{\sin B}{\sin C}$

 $\Rightarrow\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{b}{c}\left(\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\right)$

$\Rightarrow$    $b^{2}+c^{2}-a^{2}=2b^{2}$

$\Rightarrow$    $c^{2}-a^{2}=b^{2}$

$\Rightarrow$     $c^{2}=a^{2}+b^{2}$

 $\Rightarrow$   $\triangle$ ABC  right angled  at  $\angle C$