Let's solve the given problems from the image for problems 28 and 30.

Problem 28:

$m\angle ABC = 4x - 12, \quad m\angle ABD = 24$ Since $BD$ bisects $\angle ABC$, we know: $m\angle ABD = m\angle DBC$ This implies: $m\angle ABC = 2 \times m\angle ABD$ Thus, substitute the given value: $4x - 12 = 2 \times 24$ Simplify: $4x - 12 = 48$ Add 12 to both sides: $4x = 60$ Divide by 4: $x = 15$ Now substitute $x = 15$ back into the expression for $m\angle ABC$: $m\angle ABC = 4(15) - 12 = 60 - 12 = 48$ So: $m\angle ABC = 48^\circ$

Problem 30:

$m\angle ABD = 3x + 20, \quad m\angle CBD = 6x - 16$ Since $BD$ bisects $\angle ABC$, we know: $m\angle ABD = m\angle DBC$ So, equate the two expressions: $3x + 20 = 6x - 16$ Solve for $x$: $20 + 16 = 6x - 3x$ $36 = 3x$ $x = 12$ Now substitute $x = 12$ back into the expressions for $m\angle ABD$ and $m\angle DBC$: $m\angle ABD = 3(12) + 20 = 36 + 20 = 56^\circ$ $m\angle ABC = 2 \times m\angle ABD = 2 \times 56 = 112^\circ$

Summary of Results:

• Problem 28: $m\angle ABC = 48^\circ$
• Problem 30: $m\angle ABC = 112^\circ$

Let me know if you'd like further details or explanations!

Here are 5 related questions:

1. How do you verify if an angle bisector equally divides an angle?
2. What are other geometric properties of angle bisectors in a triangle?
3. How can algebraic techniques simplify angle-related problems?
4. What happens when two angles are supplementary in a bisected configuration?
5. Can you generalize this process for solving any linear equation related to angles?

Tip: Always substitute back the value of $x$ to check the correctness of your final angle calculations!