1. **Problem 1: Find angle $x$ in a regular pentagon where $x$ is the angle at vertex $B$ between sides $AB$ and diagonal $BD$.** - A regular pentagon has 5 equal interior angles and 5 equal sides. - Each interior angle of a regular pentagon is given by $$\frac{(5-2) \times 180}{5} = \frac{3 \times 180}{5} = 108^\circ.$$ - The diagonal $BD$ divides the pentagon in such a way that angle $x$ at vertex $B$ is an exterior segment angle. - To find $x$, consider triangle $ABD$. - Using the properties and symmetry of a regular pentagon and its diagonals it is known that the angle between a side and a diagonal at a vertex is $$36^\circ.$$ Thus, $$x = 36^\circ.$$ 2. **Problem 2: Find angle $PRQ$ in a regular 12-sided polygon (dodecagon) where $PQ$ and $QR$ are sides and $PR$ is a diagonal.** - The interior angle of a regular 12-sided polygon is $$\frac{(12-2) \times 180}{12} = \frac{10 \times 180}{12} =150^\circ.$$ - The central angle between adjacent vertices is $$\frac{360}{12} = 30^\circ.$$ - Triangle $PQR$ includes sides $PQ$, $QR$, and diagonal $PR$ skipping one vertex, so $PR$ subtends twice the central angle: $$2 \times 30^\circ = 60^\circ.$$ - Triangle $PQR$ is isosceles with $PQ=QR$ as sides of the polygon. - The angle at $R$, $PRQ$, can be calculated using triangle angle sum: $$\text{Angle } PRQ = \frac{180^\circ - 60^\circ}{2} = 60^\circ.$$ So, $$PRQ = 60^\circ.$$ 3. **Problem 3: Find angle $x$ where a regular hexagon and regular octagon join, and $x$ is the angle at their shared vertex.** - Interior angle of regular hexagon: $$\frac{(6-2) \times 180}{6} = 120^\circ.$$ - Interior angle of regular octagon: $$\frac{(8-2) \times 180}{8} = 135^\circ.$$ - At the shared vertex, $x$ is the external angle formed between the hexagon and octagon sides. - The sum of interior angles around a point is $360^\circ$. Therefore, $$x = 360^\circ - (120^\circ + 135^\circ) = 105^\circ.$$ 4. **Problem 4: Find angle $x$ in a rhombus inside a regular hexagon at the upper right vertex of the rhombus inside the hexagon.** - Interior angle of regular hexagon is $120^\circ$. - A rhombus has opposite equal angles. - The rhombus is inscribed sharing a vertex of the hexagon, so angle $x$ at that vertex reflects half the hexagon's angle due to symmetry and rhombus properties. Thus, $$x = 60^\circ.$$