Hey, good to see you again. Step 1: Calculate the interior angle of a regular pentagon. A regular pentagon has $n=5$ sides. The formula for the interior angle of a regular $n$-sided polygon is $\frac{(n-2) \times 180^\circ}{n}$. $$ \angle EAB = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ $$ a) $\boxed{\text{108}^\circ}$ Step 2: Calculate $\angle ABN$. In a regular pentagon, all sides are equal in length. Therefore, $EA = AB$. This means $\triangle EAB$ is an isosceles triangle. The base angles of an isosceles triangle are equal. So, $\angle ABE = \angle AEB$. The sum of angles in a triangle is $180^\circ$. $$ \angle EAB + \angle ABE + \angle AEB = 180^\circ $$ $$ 108^\circ + 2 \times \angle ABE = 180^\circ $$ $$ 2 \times \angle ABE = 180^\circ - 108^\circ $$ $$ 2 \times \angle ABE = 72^\circ $$ $$ \angle ABE = \frac{72^\circ}{2} = 36^\circ $$ Since N lies on EB, $\angle ABN = \angle ABE$. b)