Let's solve the given question step by step for each polygon. The goal is to find the missing angle $x$ in each part:

(a) Triangle

In any triangle, the sum of the angles is $180^\circ$. The given angles are $30^\circ$ and $90^\circ$. Thus: $x + 90^\circ + 30^\circ = 180^\circ$ $x = 180^\circ - 120^\circ$ $x = 60^\circ$

(b) Quadrilateral

In any quadrilateral, the sum of the angles is $360^\circ$. The given angles are $80^\circ$, $60^\circ$, and a right angle ($90^\circ$). Thus: $x + 80^\circ + 60^\circ + 90^\circ = 360^\circ$ $x = 360^\circ - (80^\circ + 60^\circ + 90^\circ)$ $x = 360^\circ - 230^\circ$ $x = 130^\circ$

(c) Pentagon

In any pentagon, the sum of the angles is $(n-2) \cdot 180^\circ$, where $n = 5$. Thus: $\text{Sum of angles} = (5-2) \cdot 180^\circ = 540^\circ$ All the other angles are $120^\circ$ each, and the last angle is $x$. Thus: $120^\circ + 120^\circ + 120^\circ + 120^\circ + x = 540^\circ$ $x = 540^\circ - 480^\circ$ $x = 60^\circ$

(d) Triangle

In any triangle, the sum of the angles is $180^\circ$. The given angles are $68^\circ$ and $120^\circ$. Thus: $x + 68^\circ + 120^\circ = 180^\circ$ $x = 180^\circ - 188^\circ$ $x = -8^\circ$ (Might want to recheck on the input values -- double-check!)

Let me clarify