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

In a $\triangle ABC$,if $b=10$,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$,and the area of the triangle is $15\sqrt{3}$ sq. units,then $\cot \frac{B}{2} =$

In $\Delta ABC,$ if $2(bc \cos A + ca \cos B + ab \cos C) = $

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$,respectively. Then which of the following statements is (are) $TRUE$?
$(A)$ $\cos P \geq 1-\frac{p^2}{2qr}$
$(B)$ $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
$(C)$ $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$
$(D)$ If $p < q$ and $p < r$,then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$

In $\triangle ABC$ with usual notation,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=\frac{1}{\sqrt{6}}$,then the area of the triangle is

The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta = \cot 5 \theta$ as well as $\sin 2 \theta = \cos 4 \theta$ is