In triangles $PQR$ and $MST$,$\angle P = 55^{\circ}$,$\angle Q = 25^{\circ}$,$\angle M = 100^{\circ}$ and $\angle S = 25^{\circ}$. Is $\triangle QPR \sim \triangle TSM$? Why?

(N/A) No,$\triangle QPR$ is not similar to $\triangle TSM$.
We know that the sum of the three angles of a triangle is $180^{\circ}$.
In $\triangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.
$55^{\circ} + 25^{\circ} + \angle R = 180^{\circ}$.
$\angle R = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
In $\triangle TSM$,$\angle T + \angle S + \angle M = 180^{\circ}$.
$\angle T + 25^{\circ} + 100^{\circ} = 180^{\circ}$.
$\angle T = 180^{\circ} - 125^{\circ} = 55^{\circ}$.
Comparing the angles of $\triangle PQR$ and $\triangle TSM$:
$\angle P = 55^{\circ} = \angle T$
$\angle Q = 25^{\circ} = \angle S$
$\angle R = 100^{\circ} = \angle M$
Since all corresponding angles are equal,the triangles are similar,but the correct correspondence is $\triangle PQR \sim \triangle TSM$.
Therefore,$\triangle QPR$ is not similar to $\triangle TSM$ because the order of vertices does not match the correspondence of equal angles.