$\angle A = \angle B$. If $\angle A$ and $\angle B$ are supplementary angles,then $\angle A = \ldots$ (in $^o$)

If one of the two angles of a linear pair has measure $x^{\circ},$ then the other angle has measure of $\ldots \ldots . .$

If one angle of a triangle is equal to the sum of the other two angles,then the triangle is

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2:3,$ then the greater of the two angles is (in $^{\circ}$)

Bisectors of angles $B$ and $C$ of a triangle $ABC$ intersect each other at the point $O$. Prove that $\angle BOC = 90^{\circ} + \frac{1}{2} \angle A$.

Can a triangle have all angles less than $60^{\circ}$? Give reason for your answer.