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For problem 42, we are given:

- \( m \angle EBC = 3r + 10 \)
- \( m \angle ABE = 2r - 20 \)

Since \( BH \) bisects \( \angle EBC \), we know:

\[
m \angle EBF = m \angle EBC - m \angle ABE
\]

Setting up the equation and solving, we find that:

- \( r = -30 \)

Substituting \( r = -30 \) into the equation for \( m \angle EBF \), we find that:

\[
m \angle EBF = 0^\circ
\]

Would you like to see more details about the solution process or ask any further questions?

Here are some related questions you might explore:
1. How do angle bisectors work in geometry?
2. Why is \( m \angle EBF \) equal to 0 degrees in this case?
3. Can you explain the significance of opposite rays in this problem?
4. What happens if \( r \) is positive? Is the setup always consistent?
5. How does the equation change if the angle bisector condition is modified?

**Tip**: Always double-check the logic of geometric constraints when solving equations involving bisected angles.

Resolve problem 42: If m∠EBC = 3r + 10 and m∠ABE = 2r - 20, find m∠EBF.

In this geometry problem, learn how to solve for m∠EBF by using the angle bisector theorem and linear equations. Given m∠EBC = 3r + 10 and m∠ABE = 2r - 20, follow a step-by-step guide to find the value of r and calculate the angle. Suitable for students in Grades 9-10.

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