1. **State the problem:** In triangle $PQR$, given $\angle Q = 50^\circ$ and $\angle R = 40^\circ$, $PS$ bisects $\angle QPR$, and $PT \perp QR$. Find the value of $\angle TPS + 2\angle PST + 2\angle QPT$. 2. **Find $\angle P$:** The sum of angles in a triangle is $180^\circ$. So, $$\angle P = 180^\circ - \angle Q - \angle R = 180^\circ - 50^\circ - 40^\circ = 90^\circ.$$ 3. **Interpret the given information:** Since $PS$ bisects $\angle QPR$, it divides $\angle P$ into two equal angles of $45^\circ$ each. 4. **Analyze $PT \perp QR$:** $PT$ is perpendicular to $QR$, so $\angle QPT = 90^\circ$. 5. **Identify angles:** - $\angle QPT = 90^\circ$ (given by perpendicularity). - $\angle QPR = 90^\circ$ (from step 2). - $PS$ bisects $\angle QPR$, so $\angle QPS = \angle SPR = 45^\circ$. 6. **Find $\angle PST$ and $\angle TPS$:** Since $PT$ is perpendicular to $QR$, and $S$ lies on $QR$, triangle $PST$ is right-angled at $T$. 7. **Calculate the sum:** $$\angle TPS + 2\angle PST + 2\angle QPT = ?$$ Since $\angle QPT = 90^\circ$, $$2\angle QPT = 2 \times 90^\circ = 180^\circ.$$ $\angle PST$ is half of $\angle SPR = 45^\circ$ because $PS$ bisects $\angle QPR$ and $T$ lies on $QR$, so $\angle PST = 45^\circ$. $\angle TPS$ is complementary to $\angle PST$ in triangle $PST$ (right angled at $T$), so $$\angle TPS = 90^\circ - 45^\circ = 45^\circ.$$ 8. **Sum all:** $$\angle TPS + 2\angle PST + 2\angle QPT = 45^\circ + 2 \times 45^\circ + 180^\circ = 45^\circ + 90^\circ + 180^\circ = 315^\circ.$$ **Final answer:** $$\boxed{315^\circ}.$$