The measures of two vertically opposite angles are $(5x + 30^{\circ})$ and $(8x - 60^{\circ})$,then the value of $x = \dots$

$AP$ and $BQ$ are the bisectors of the two alternate interior angles formed by the intersection of a transversal $t$ with parallel lines $l$ and $m$ (see figure). Show that $AP \parallel BQ$.

If one of the angles formed by two intersecting lines is a right angle,what can you say about the other three angles? Give reason for your answer.

The supplementary angle of the complementary angle of an angle having measure $32^{\circ}$ has measure.......... (in $^{\circ}$)

If two parallel lines are intersected by a transversal,then interior angles on the same side of the transversal are $\ldots \ldots . . .$

In $\Delta ABC$,the sides $AB$ and $AC$ are produced to $D$ and $E$ respectively,so that exterior angles $\angle CBD$ and $\angle BCE$ are formed. If bisectors of $\angle CBD$ and $\angle BCE$ intersect at point $O$,prove that $\angle BOC = 90^{\circ} - \frac{1}{2} \angle A$.