In a triangle $$\mathrm{ABC}$$, with usual notations, if $$\mathrm{c}=4$$, then value of $$(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}$$ is

If one side of a triangle is double the other and the angles opposite to these sides differ by $$60^{\circ}$$, then the triangle is

In $$\triangle \mathrm{PQR}, \sin \mathrm{P}, \sin \mathrm{Q}$$ and $$\sin \mathrm{R}$$ are in A.P., then

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 $$\frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}=$$