The problem involves a regular 8-sided polygon (octagon) and a regular 10-sided polygon (decagon) sharing a common side, and we are asked to calculate the size of the angle marked $b$ at the intersection of these polygons.

Step 1: Calculate the interior angles of the polygons

For any regular polygon with $n$ sides, the formula for the interior angle $\theta$ is given by:

$\theta = \frac{(n-2) \times 180^\circ}{n}$

• For the octagon (8-sided polygon):

$\theta_{\text{octagon}} = \frac{(8-2) \times 180^\circ}{8} = \frac{6 \times 180^\circ}{8} = 135^\circ$

• For the decagon (10-sided polygon):

$\theta_{\text{decagon}} = \frac{(10-2) \times 180^\circ}{10} = \frac{8 \times 180^\circ}{10} = 144^\circ$

Step 2: Calculate the exterior angles of the polygons

The exterior angle $\phi$ of a polygon is related to its interior angle by:

$\phi = 180^\circ - \theta$

• For the octagon:

$\phi_{\text{octagon}} = 180^\circ - 135^\circ = 45^\circ$

• For the decagon:

$\phi_{\text{decagon}} = 180^\circ - 144^\circ = 36^\circ$

Step 3: Angle at point $b$

At the vertex where the two polygons meet, the exterior angles combine to form a straight line. The sum of these angles must be 180° since they lie along a straight line. Hence, the angle $b$ is:

$b = 180^\circ - (\phi_{\text{octagon}} + \phi_{\text{decagon}})$

$b = 180^\circ - (45^\circ + 36^\circ) = 180^\circ - 81^\circ = 99^\circ$

Final Answer:

The size of the angle marked $b$ is $\boxed{99^\circ}$.

Let me know if you'd like more details, or if you have any questions!

Here are 5 questions that expand on this concept:

1. How would the calculation change if one polygon were a pentagon instead of a decagon?
2. What is the general formula for the exterior angle of a regular polygon?
3. Can the angle $b$ ever be less than $90^\circ$ for any regular polygons?
4. How can you determine the sum of interior angles of a polygon without calculating each individual angle?
5. What happens to the angle $b$ as the number of sides of one of the polygons increases?

Tip: Always remember that the sum of exterior angles of any polygon is always $360^\circ$, no matter how many sides it has!