1. **Stating the problem:** We have triangle ABC with point D on segment AC such that ADC is a straight line. Given angles: $\angle A = 77^\circ$, $\angle C = 35^\circ$, $\angle D = y^\circ$, and two angles labeled $x^\circ$ inside the triangle at vertex B. We need to find $x$ and $y$.\n\n2. **Key fact:** Since ADC is a straight line, $\angle ADC = 180^\circ$.\n\n3. **Sum of angles in triangle ABC:** $\angle A + \angle B + \angle C = 180^\circ$. Here, $\angle B = 2x$ because there are two angles labeled $x^\circ$ at B.\n\n4. Write the equation for triangle ABC:\n$$77^\circ + 2x + 35^\circ = 180^\circ$$\n\n5. Simplify:\n$$112^\circ + 2x = 180^\circ$$\n\n6. Solve for $x$:\n$$2x = 180^\circ - 112^\circ$$\n$$2x = 68^\circ$$\n$$x = \frac{68^\circ}{2}$$\n$$x = 34^\circ$$\n\n7. Since ADC is a straight line, angles $\angle BDC = y^\circ$ and $\angle BDA = x^\circ + 35^\circ$ must sum to 180° (linear pair). But here, $y$ is the angle at D between B and C, so:\n$$y + 77^\circ = 180^\circ$$\n\n8. Solve for $y$:\n$$y = 180^\circ - 77^\circ$$\n$$y = 103^\circ$$\n\n**Final answers:**\n$$x = 34^\circ, \quad y = 103^\circ$$