In a $\Delta ABC$,the sides $a$,$b$,$c$ are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$; then $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

The upper $\frac{3}{4}$th portion of a vertical pole subtends an angle $\tan^{-1}\left(\frac{3}{5}\right)$ at a point in the horizontal plane through its foot and at a distance $40 \ m$ from the foot. $A$ possible height of the vertical pole is $....... \ m$.

With usual notations in $\triangle ABC$,if $\angle B = \frac{\pi}{2}$,and $\tan \frac{A}{2}, \tan \frac{C}{2}$ are roots of the equation $px^2 + qx + r = 0$,$p \neq 0$,then:

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) = $

If ${c^2} = {a^2} + {b^2}$,then $4s(s - a)(s - b)(s - c) = $

If $a \cos A = b \cos B$,then $\Delta ABC$ is