Let $$a, b, c$$ be the lengths of sides of triangle $$A B C$$ such that $$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$$. Then $$\frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}=$$

In $$\triangle \mathrm{ABC}$$, with usual notations, $$\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$$, if $$\tan \left(\frac{A}{2}\right)$$ and $$\tan \left(\frac{B}{2}\right)$$ are the roots of the equation $$a_1 x^2+b_1 x+c_1=0\left(a_1 \neq 0\right)$$, then

Two sides of a triangle are $$\sqrt{3}+1$$ and $$\sqrt{3}-1$$ and the included angle is $$60^{\circ}$$, then the difference of the remaining angles is

In a triangle ABC with usual notations a = 2, b = 3, then value of $$\frac{\cos 2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^2}$$ is