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

In a triangle $ABC$,$BC=7$,$AC=8$,$AB=\alpha \in N$ and $\cos A=\frac{2}{3}$. If $49 \cos (3C)+42=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to..........

If the equation $\sin^4 x - (p+2) \sin^2 x - (p+3) = 0$ has a solution, then $p$ must lie in the interval:

If $ABC$ is a triangle of area $\Delta$ with $a = 2$, $b = \frac{7}{2}$, $c = \frac{5}{2}$, where $a, b, c$ are the lengths of the sides of the triangle opposite to angles $A, B$ and $C$ respectively, then $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A}$ is equal to...

$\alpha, \beta$ are the roots of the equation $\sin^2 x + b \sin x + c = 0$. If $\alpha + \beta = \frac{\pi}{2}$,then $b^2 - 1 =$

In a triangle $ABC$,evaluate the expression $\frac{a(r_1 r + r_2 r_3)}{r_1 - r + r_2 + r_3}$.