In $\Delta ABC$,$AB = 8 \, cm$ and $BC = 5 \, cm$,then $AC > \ldots \ldots \ldots cm$.

"If two angles and a side of one triangle are equal to two angles and a side of another triangle,then the two triangles must be congruent." Is the statement true? Why?

In $\Delta ABC$ and $\Delta PQR$,if $\angle B = \angle P = 90^{\circ}$,$AC = RQ$,and $AB = RP$,then $\Delta ABC \cong \Delta \ldots \ldots \ldots$

Is it possible to construct a triangle with lengths of its sides as $4 \, cm, 3 \, cm$ and $7 \, cm$? Give reason for your answer.

In $\Delta ABC$,$AB = AC$ and $\angle C = 75^{\circ}$,then find $\angle B$. (in $^{\circ}$)

Bisectors of the angles $B$ and $C$ of an isosceles triangle $ABC$ with $AB = AC$ intersect each other at $O$. Show that the external angle adjacent to $\angle ABC$ is equal to $\angle BOC$.