In the figure, what is the value of $\angle \mathrm{r}-\angle \mathrm{p}$ if $\angle \mathrm{q}={140}^{\mathrm{o}}$ ?

As shown in the figure, angles p and q form a linear pair.

Because $\angle \mathrm{q}=140°,\angle \mathrm{p}=180°-40°=40°$
In addition, the angle in the right-angled triangle opposite p will be equal to p = 40^o, because $\angle \mathrm{r}$ is the exterior angle of the given right-angled triangle,
$\mathrm{r}=90°+40°=130°$
$\therefore \angle \mathrm{r}-\angle \mathrm{p}=90°$