In triangles \(PQR\) and \(MST\), \(\angle P=55^\circ,\; \angle Q=25^\circ\) and \(\angle M=100^\circ,\; \angle S=25^\circ\). Is \(\triangle QPR \sim \triangle TSM\)? Why?
No. But \(\triangle QPR \sim \triangle STM\).
Step 1: Find the missing angle in triangle PQR.
We know the sum of angles in any triangle is \(180^\circ\).
In \(\triangle PQR\):
\(\angle R = 180^\circ - (\angle P + \angle Q)\)
\(= 180^\circ - (55^\circ + 25^\circ)\)
\(= 180^\circ - 80^\circ = 100^\circ\).
So the three angles of \(\triangle PQR\) are: 55°, 25°, and 100°.
Step 2: Find the missing angle in triangle MST.
In \(\triangle MST\):
\(\angle T = 180^\circ - (\angle M + \angle S)\)
\(= 180^\circ - (100^\circ + 25^\circ)\)
\(= 180^\circ - 125^\circ = 55^\circ\).
So the three angles of \(\triangle MST\) are: 100°, 25°, and 55°.
Step 3: Match the equal angles of both triangles.
Step 4: Write the correct order of similarity.
Corresponding equal angles must come in the same order.
So the correct similarity is: \(\triangle QPR \sim \triangle STM\).
Not \(\triangle TSM\).