Let's analyze the triangle in the diagram and solve for $k$ as requested in part (a).

Given:

• $\angle ACB = 80^\circ$
• $\angle ABC = (8k + 20^\circ)$
• $\angle BDA = (3k + 15^\circ)$

Step 1: Use the property of angles in a triangle

In any triangle, the sum of interior angles is always $180^\circ$. For triangle $ABC$: $\angle ACB + \angle ABC + \angle BAC = 180^\circ$ Let $\angle BAC = (3k + 15^\circ)$, as it appears $\angle BDA$ is an extension of $\angle BAC$.

Now, we substitute the values: $80^\circ + (8k + 20^\circ) + (3k + 15^\circ) = 180^\circ$

Step 2: Simplify the equation

$80^\circ + 8k + 20^\circ + 3k + 15^\circ = 180^\circ$ $115^\circ + 11k = 180^\circ$

Step 3: Solve for $k$

Subtract $115^\circ$ from both sides: $11k = 65^\circ$ Now, divide by 11: $k = \frac{65}{11} \approx 5.91^\circ$

Part (a) Answer:

The value of $k$ is approximately $5.91^\circ$.

Now, for part (b), finding the size of $\angle CBD$.

This is a vertically opposite angle to $\angle ABC$, so: $\angle CBD = \angle ABC = 8k + 20^\circ$

Substitute $k = 5.91$: $\angle CBD = 8(5.91) + 20 = 47.28 + 20 = 67.28^\circ$

Part (b) Answer:

The size of $\angle CBD$ is approximately $67.28^\circ$.

Do you need more details or clarification on any step?

Related questions:

1. How do you find angles using the exterior angle theorem?
2. What are vertically opposite angles and how do you use them in problems?
3. How do you solve for unknowns in an equation involving angles?
4. What are the sum of angles in different types of polygons?
5. How does the angle sum property change in non-Euclidean geometry?

Tip:

Always remember that in any triangle, the sum of the interior angles is always $180^\circ$. This can help you solve for unknown angles effectively!