Let’s analyze and solve the problems from the uploaded image. Below are detailed solutions for each question:

Problem 8:

Given:

• $\angle BAC = 80^\circ$
• $\overrightarrow{AM}$ and $\overrightarrow{AN}$ are interior bisectors.
• Ratio of $\angle MAN : \angle BAN = 7 : 4$.

We are tasked to find $\angle MAN$.

Solution: Let the measure of $\angle MAN$ be $7x$ and $\angle BAN$ be $4x$. Since these two angles are interior subdivisions of $\angle BAC = 80^\circ$, their sum is: $7x + 4x = 80^\circ$

$11x = 80^\circ \implies x = \frac{80^\circ}{11}$

Now, calculate $\angle MAN = 7x$: $\angle MAN = 7 \cdot \frac{80^\circ}{11} = \frac{560^\circ}{11} \approx 50.91^\circ \approx 51^\circ.$

However, since the provided answer options do not include this exact number, I may refine the calculation if the ratio has additional context.

Answer (Tentative): Approx $51^\circ$. Please confirm if more clarification is needed.

Problem 9:

Given:

• $\overrightarrow{BP}$ and $\overrightarrow{BQ}$ are bisectors of $\angle ABD$ and $\angle CBD$, respectively.
• $\angle ABC = 140^\circ$. We need to find $\angle PBQ$.

Solution: The bisectors divide the angles $\angle ABD$ and $\angle CBD$ into equal parts. Hence: $\angle ABD + \angle CBD = \angle ABC = 140^\circ.$

Since $\overrightarrow{BP}$ and $\overrightarrow{BQ}$ are bisectors: $\angle ABP = \frac{\angle ABD}{2}, \quad \angle CBQ = \frac{\angle CBD}{2}.$

Let $\angle ABD = x$ and $\angle CBD = 140^\circ - x$.

The angle between the two bisectors is: $\angle PBQ = \frac{x}{2} + \frac{140^\circ - x}{2} = \frac{140^\circ}{2} = 70^\circ.$

Answer: $\angle PBQ = 70^\circ$.

Problem 10:

Given:

• $\angle AOB = 110^\circ$,
• $\angle POM = 60^\circ$,
• $\angle AOM = \angle BOP$.

We need to find $\angle POV$.

Solution: The key is to recognize the relationships between the given angles.

1. $\angle AOB = \angle AOM + \angle BOP = 110^\circ$. Since $\angle AOM = \angle BOP$, we can write: $2 \cdot \angle AOM = 110^\circ \implies \angle AOM = 55^\circ.$

2. $\angle POM = 60^\circ$, so $\angle POV = \angle POM + \angle BOP$: $\angle POV = 60^\circ + 55^\circ = 115^\circ.$

Answer: $\angle POV = 115^\circ$.

Do you want more clarification or step-by-step details for any of these problems? Here are five related questions to deepen understanding:

1. Problem 8: How can we verify if $\angle MAN$ has been calculated correctly with respect to the ratio $7:4$?
2. Problem 9: What happens to $\angle PBQ$ if $\angle ABC$ is less than $140^\circ$?
3. Problem 10: Can $\angle POV$ exceed $180^\circ$ if different initial conditions are provided?
4. How can you generalize the bisector property in Problem 9 to a case with $n$-division of angles?
5. How would the solutions change if the drawings were not to scale or had inconsistent labeling?

Tip: For angle bisector problems, always verify that the sum of divided angles equals the original angle, as this ensures the calculations align with the geometric constraints.