Question: Solve 38 and 40
1. **Problem Statement:**
Solve for $x$ and find the measures of the angles in problems 38 and 40 where $BD$ bisects $\angle ABC$.
2. **Given:**
- Problem 38: Angles are $(3x + 6)^\circ$ and $(7x - 18)^\circ$.
- Problem 40: Angles are $(8x + 35)^\circ$ and $(11x + 23)^\circ$.
Since $BD$ bisects $\angle ABC$, the two angles are congruent.
3. **Formula:**
If $BD$ bisects $\angle ABC$, then
$$m\angle ABD = m\angle CBD$$
4. **Solve Problem 38:**
Set the angles equal:
$$3x + 6 = 7x - 18$$
Subtract $3x$ from both sides:
$$\cancel{3x} + 6 = \cancel{3x} + 7x - 18 \implies 6 = 4x - 18$$
Add $18$ to both sides:
$$6 + 18 = 4x - 18 + 18 \implies 24 = 4x$$
Divide both sides by $4$:
$$\frac{24}{\cancel{4}} = \frac{4x}{\cancel{4}} \implies 6 = x$$
Find $m\angle ABD$:
$$3x + 6 = 3(6) + 6 = 18 + 6 = 24^\circ$$
Find $m\angle CBD$:
$$7x - 18 = 7(6) - 18 = 42 - 18 = 24^\circ$$
Find $m\angle ABC$ (sum of both angles):
$$24^\circ + 24^\circ = 48^\circ$$
5. **Solve Problem 40:**
Set the angles equal:
$$8x + 35 = 11x + 23$$
Subtract $8x$ from both sides:
$$\cancel{8x} + 35 = \cancel{8x} + 11x + 23 \implies 35 = 3x + 23$$
Subtract $23$ from both sides:
$$35 - 23 = 3x + 23 - 23 \implies 12 = 3x$$
Divide both sides by $3$:
$$\frac{12}{\cancel{3}} = \frac{3x}{\cancel{3}} \implies 4 = x$$
Find $m\angle ABD$:
$$8x + 35 = 8(4) + 35 = 32 + 35 = 67^\circ$$
Find $m\angle CBD$:
$$11x + 23 = 11(4) + 23 = 44 + 23 = 67^\circ$$
Find $m\angle ABC$ (sum of both angles):
$$67^\circ + 67^\circ = 134^\circ$$
**Final answers:**
- Problem 38: $x=6$, $m\angle ABD = 24^\circ$, $m\angle CBD = 24^\circ$, $m\angle ABC = 48^\circ$
- Problem 40: $x=4$, $m\angle ABD = 67^\circ$, $m\angle CBD = 67^\circ$, $m\angle ABC = 134^\circ$