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

With usual notations, perimeter of a triangle $$A B C$$ is 6 times the arithmetic mean of sine of its angles. If $$\mathrm{a}=1$$, then measure of angle $$\mathrm{A}=$$