The angles of a triangle are in the ratio $1:3:5$. Find the greatest angle. (in $\pi /9$)

In a triangle $ABC$,with usual notations,if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$,then $\cos A : \cos B : \cos C =$

In a triangle $ABC$,if $a=3, b=7, c=8$,then $\sin \frac{B}{2} \tan \frac{C-A}{2} =$

If in a $\Delta ABC$,$\cos 3A + \cos 3B + \cos 3C = 1$,then one angle must be exactly equal to .......$^o$

In $\Delta ABC,$ $a\sin (B - C) + b\sin (C - A) + c\sin (A - B) = $

In a triangle $ABC$, with the usual notations, $\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$. If $D$ divides $BC$ internally in the ratio $1:3$, then $\frac{\sin \angle BAD}{\sin \angle CAD} =$