With usual notation in a $\Delta ABC$,$\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \left( \frac{1}{r_2} + \frac{1}{r_3} \right) \left( \frac{1}{r_3} + \frac{1}{r_1} \right) = \frac{K R^3}{a^2 b^2 c^2}$ where $K$ has the value equal to

In $\triangle ABC$,$AD$ and $BE$ are medians drawn from $A$ and $B$. If $AD = \frac{7}{2}$,$\angle DAB = \frac{\pi}{8}$ and $\angle ABE = \frac{\pi}{4}$,then the area (in sq. units) of $\triangle ABC$ is

The angles $A, B$ and $C$ of a triangle $ABC$ are in $A.P.$ and $a : b = 1 : \sqrt{3}$. If $c = 4 \text{ cm}$,then the area (in $\text{sq. cm}$) of this triangle is:

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

Let $P_1, P_2, P_3$ be the altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively. If $\Delta$ denotes the area of the triangle and $s$ is the semi-perimeter of the triangle, then $\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} = $

In $\triangle ABC$,with usual notations,if $a, b, c$ are in $A.P.$,then $a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right) = $