If $x, y$ and $z$ are the distances of the incentre from the vertices $A, B$ and $C$ of the triangle $ABC$ respectively,then $\frac{abc}{xyz}$ is equal to

With usual notations,in $\triangle ABC$,if $a=2, b=3, c=5$ and $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$,then $k=$

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

If in a triangle $ABC$,$a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$,then its sides will be in

If the incircle of the $\Delta ABC$ touches its sides at $L, M$ and $N$ respectively,and if $x, y, z$ are the circumradii of the triangles $\Delta MIN, \Delta NIL$ and $\Delta LIM$ respectively,where $I$ is the incentre,then the product $xyz$ is equal to:

In a right-angled triangle,if the difference between the two acute angles is $60^{\circ}$,then the ratio of the length of the hypotenuse to the length of the perpendicular drawn to the hypotenuse from its opposite vertex is: (in $: 1$)