With usual notation, in triangle ABC , $\mathrm{m} \angle \mathrm{A}=30^{\circ}$ then the value of $\left(1+\frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}}\right)\left(1+\frac{\mathrm{c}}{\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{b}}\right)$ is equal to

In $\triangle A B C$, with usual notations, $a \cos B=b \cos A, a \cos C \neq c \cos A$ then $\mathrm{A}(\triangle \mathrm{ABC})$ $\qquad$ sq. units.

In a triangle ABC , with usual notations if $\mathrm{a}=4, \mathrm{~b}=8, \angle \mathrm{C}=60^{\circ}$, then the value of $\angle \mathrm{B}$ and the ratio $\cos \mathrm{A}: \cos \mathrm{C}$ respectively are,

In a triangle ABC with usual notations if $|\overline{\mathrm{BC}}|=8,|\overline{\mathrm{CA}}|=7,|\overline{\mathrm{AB}}|=10$ then the projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{AC}}$ is