1. **Problem statement:** (a) Find the measure of angle $x$ in a regular pentagon where three angles are labeled $x$. (b) Find the measure of angle $x$ in a trapezoid-like polygon with all sides equal and diagonals intersecting to form angles $120^\circ$, $60^\circ$, and $x$. --- 2. **Formulas and rules:** - The sum of interior angles of an $n$-sided polygon is given by: $$\text{Sum} = (n-2) \times 180^\circ$$ - In a regular polygon, all interior angles are equal. - When two lines intersect, the opposite angles are equal. - The sum of angles around a point is $360^\circ$. --- 3. **Solution for (a):** - A pentagon has $n=5$ sides. - Sum of interior angles: $$ (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ $$ - Since the pentagon is regular, all interior angles are equal: $$ x = \frac{540^\circ}{5} = 108^\circ $$ - Therefore, each angle $x$ measures $108^\circ$. --- 4. **Solution for (b):** - The polygon has all sides equal and two diagonals intersecting forming angles $120^\circ$, $60^\circ$, and $x$. - At the intersection of diagonals, the angles around the point sum to $360^\circ$. - The diagonals form four angles; two are given as $120^\circ$ and $60^\circ$, and one is $x$. - The opposite angle to $120^\circ$ is also $120^\circ$ (vertical angles). - The opposite angle to $60^\circ$ is also $60^\circ$. - The remaining angle $x$ is opposite to another angle $x$. - Sum of all four angles at intersection: $$ 120^\circ + 60^\circ + x + x = 360^\circ $$ - Simplify: $$ 180^\circ + 2x = 360^\circ $$ - Solve for $x$: $$ 2x = 360^\circ - 180^\circ = 180^\circ $$ $$ x = \frac{180^\circ}{2} = 90^\circ $$ - Therefore, angle $x$ measures $90^\circ$. --- **Final answers:** - (a) $x = 108^\circ$ - (b) $x = 90^\circ$