1. **Stating the problem:** We are given an equilateral triangle PQT and a quadrilateral T-S-R-Q with given angles 140° at P, 80° at S, and 130° at R. We need to find the value of angle $x^\circ$ at vertex Q between QR and QT. 2. **Important properties:** - In an equilateral triangle, all angles are equal to 60°. - The sum of angles around point Q is 360°. - The sum of interior angles in a quadrilateral is 360°. 3. **Known angles:** - $\angle P = 140^\circ$ - $\angle S = 80^\circ$ - $\angle R = 130^\circ$ - $\angle T$ in triangle PQT is 60° (equilateral triangle) 4. **Find $\angle Q$ in quadrilateral T-S-R-Q:** Sum of angles in quadrilateral = 360° $$\angle T + \angle S + \angle R + \angle Q = 360^\circ$$ $$60 + 80 + 130 + \angle Q = 360$$ $$270 + \angle Q = 360$$ $$\angle Q = 360 - 270 = 90^\circ$$ 5. **Find $x$ at vertex Q:** At point Q, the angle between QR and QT is $x^\circ$. Since $\angle Q$ in the quadrilateral is 90°, and the angle between PQ and QT in the triangle is 60°, the angle $x$ is the difference between these angles. 6. **Calculate $x$:** $$x = 90^\circ - 60^\circ = 30^\circ$$ However, the problem states $x$ is the angle between QR and QT at Q, which is the angle inside the quadrilateral, so $x = 90^\circ$. **Final answer:** $$x = 90^\circ$$ **Answer choice:** D 90°