In a triangle ABC , the sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are such that they are the roots of the equation $x^3-11 x^2+38 x-40=0$ Then

$$ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}= $$

In a triangle ABC with usual notations if $\mathrm{a}=13$, $b=14, c=15$ Then $\sin A=$

In a triangle $A B C$, with usual notations, $3 \mathrm{~b}=\mathrm{a}+\mathrm{c}$, then $\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{C}}{2}=$

In a triangle ABC , with usual notations if $\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}$ then $\angle \mathrm{A}=$