In a $\triangle ABC$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1, r_2$, and $r_3$ are

If $\triangle ABC$ is a non-isosceles triangle and $\angle C = 90^{\circ}$,then $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = $

If in a $\triangle ABC$, $r_1 < r_2 < r_3$, then:

Let $ABC$ be a triangle such that $\angle ACB = \frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value$(s)$ of $x$ for which $a = x^2+x+1, b = x^2-1$ and $c = 2x+1$ is (are)

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\frac{\cos A + 3 \cos C}{\cos B} = $

The ratios of the sides of a triangle $ABC$ are $5:12:13$ and its area is $270$. Then the sides of the triangle are: