6. if (mangle cdf=(3x + 14)^{circ}), (mangle fde=(5x - 2)^{circ}), and (mangle cde=(10x - 18)^{circ}), find each measure.
7. if (mangle lmp) is 11 degrees more than (mangle nmp) and (mangle nml = 137^{circ}), find each measure.
8. if (mangle abc) is one degree less than three times (mangle abd) and (mangle dbc = 47^{circ}), find each measure.
9. if (overline{qs}) bisects (angle pqt), (mangle sqt=(8x - 25)^{circ}), (mangle pqt=(9x + 34)^{circ}), and (mangle sqr = 112^{circ}), find each measure.
10. if (angle cde) is a straight angle, (overline{de}) bisects (angle gdh), (mangle gde=(8x - 1)^{circ}), (mangle edh=(6x + 15)^{circ}), and (mangle cdf = 43^{circ}), find each measure.

Question

Accepted Answer

### Problem 6
# Explanation:
## Step1: Use angle - addition postulate
Since $\angle CDF+\angle FDE=\angle CDE$, we have the equation $(3x + 14)+(5x - 2)=10x-18$.
## Step2: Simplify the left - hand side of the equation
Combining like terms, $3x+5x + 14-2=8x + 12$. So, $8x + 12=10x-18$.
## Step3: Solve for $x$
Subtract $8x$ from both sides: $12 = 10x-8x-18$, which simplifies to $12 = 2x-18$. Then add 18 to both sides: $12 + 18=2x$, so $30 = 2x$. Divide both sides by 2, and we get $x = 15$.
## Step4: Find the measure of each angle
$m\angle CDF=3x + 14=3	imes15+14=45 + 14=59^{\circ}$.
$m\angle FDE=5x - 2=5	imes15-2=75 - 2=73^{\circ}$.
$m\angle CDE=10x-18=10	imes15-18=150 - 18=132^{\circ}$.

# Answer:
$x = 15$
$m\angle CDF=59^{\circ}$
$m\angle FDE=73^{\circ}$
$m\angle CDE=132^{\circ}$

### Problem 7
Let $m\angle NMP=x$. Then $m\angle LMP=x + 11$.
Since $\angle LMP+\angle NMP=\angle NML$ and $\angle NML = 137^{\circ}$, we have the equation $x+(x + 11)=137$.
Combining like terms gives $2x+11 = 137$.
Subtract 11 from both sides: $2x=137 - 11=126$.
Divide both sides by 2: $x = 63$.
So, $m\angle NMP=63^{\circ}$ and $m\angle LMP=63 + 11=74^{\circ}$.

# Answer:
$m\angle LMP=74^{\circ}$
$m\angle NMP=63^{\circ}$

### Problem 8
Let $m\angle ABD=x$. Then $m\angle ABC=3x-1$.
Since $\angle ABC=\angle ABD+\angle DBC$ and $\angle DBC = 47^{\circ}$, we have the equation $3x-1=x + 47$.
Subtract $x$ from both sides: $3x-x-1=47$, which simplifies to $2x-1=47$.
Add 1 to both sides: $2x=47 + 1=48$.
Divide both sides by 2: $x = 24$.
So, $m\angle ABD=24^{\circ}$ and $m\angle ABC=3	imes24-1=72 - 1=71^{\circ}$.

# Answer:
$m\angle ABD=24^{\circ}$
$m\angle ABC=71^{\circ}$

### Problem 9
Since $\overrightarrow{QS}$ bisects $\angle PQT$, $m\angle PQS=m\angle SQT$ and $m\angle PQT = 2m\angle SQT$.
We know that $m\angle SQT=8x - 25$ and $m\angle PQT=9x + 34$. Also, $m\angle PQT = 2m\angle SQT$, so $9x+34=2(8x - 25)$.
Expand the right - hand side: $9x+34=16x-50$.
Subtract $9x$ from both sides: $34=16x-9x-50$, which simplifies to $34 = 7x-50$.
Add 50 to both sides: $34 + 50=7x$, so $84 = 7x$.
Divide both sides by 7: $x = 12$.
$m\angle PQS=m\angle SQT=8x - 25=8	imes12-25=96 - 25=71^{\circ}$.
$m\angle PQT=9x + 34=9	imes12+34=108 + 34=142^{\circ}$.
$m\angle TQR=m\angle SQR-m\angle SQT=112-(8x - 25)=112-(8	imes12 - 25)=112-(96 - 25)=112 - 71=41^{\circ}$.

# Answer:
$x = 12$
$m\angle PQS=71^{\circ}$
$m\angle PQT=142^{\circ}$
$m\angle TQR=41^{\circ}$

### Problem 10
Since $\overrightarrow{DE}$ bisects $\angle GDH$, $m\angle GDE=m\angle EDH$.
So, $8x-1=6x + 15$.
Subtract $6x$ from both sides: $8x-6x-1=15$, which simplifies to $2x-1=15$.
Add 1 to both sides: $2x=15 + 1=16$.
Divide both sides by 2: $x = 8$.
$m\angle GDH=2m\angle GDE=2(8x - 1)=2(8	imes8 - 1)=2(64 - 1)=126^{\circ}$.
Since $\angle CDE$ is a straight angle ($180^{\circ}$) and $m\angle CDF = 43^{\circ}$, $m\angle FDE=180 - 43=137^{\circ}$.
$m\angle FDH=m\angle FDE+m\angle EDH$. First, $m\angle EDH=6x + 15=6	imes8+15=48 + 15=63^{\circ}$. So, $m\angle FDH=137+63=200^{\circ}$ (but angles are usually considered between $0^{\circ}$ and $360^{\circ}$, and we might be interested in the non - reflex angle, if we consider the non - overlapping parts, we need to adjust based on the context. Assuming we are working in the standard non - reflex sense, we note that the sum of angles around a point is $360^{\circ}$, and we can also calculate in a more appropriate way considering the given information in a non - overlapping sense. Since $\angle CDF = 43^{\circ}$ and $\angle GDH=126^{\circ}$, and $\angle CDE = 180^{\circ}$, we can find the relevant non - reflex angle. The non - reflex $m\angle FDH=360-(43 + 126)=191^{\circ}$ (again, depending on the context of angle measurement, if we consider the non - overlapping part of the angles formed by the rays, we have $m\angle FDH = 191^{\circ}$). A more standard way considering the non - reflex angle formed by the rays in the figure: $m\angle FDH=180 - 43+63=200^{\circ}$ (but if we want the non - reflex angle within $0^{\circ}$ and $180^{\circ}$, we need to re - evaluate based on the actual geometric situation. Let's assume we want the non - reflex angle formed by the relevant rays in the non - overlapping sense. $m\angle FDH = 191^{\circ}$ is a reflex angle. If we consider the non - reflex part, we know that $\angle CDF = 43^{\circ}$ and $\angle GDH = 126^{\circ}$ and $\angle CDE=180^{\circ}$. The non - reflex $m\angle FDH=180 - 43+63=200^{\circ}$ (if we consider the sum of non - overlapping angles formed by the rays). In a more geometrically appropriate non - reflex sense, we note that $\angle CDF = 43^{\circ}$, $\angle GDE=\angle EDH = 63^{\circ}$, and $\angle CDE = 180^{\circ}$. The non - reflex $m\angle FDH=180-43 + 63=200^{\circ}$ (reflex), the non - reflex angle we might be interested in (formed by non - overlapping rays) is $360-200 = 160^{\circ}$).

# Answer:
$x = 8$
$m\angle GDH=126^{\circ}$
$m\angle FDH=160^{\circ}$ (assuming non - reflex angle formed by relevant non - overlapping rays in the geometric figure)

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QUESTION IMAGE

Question

Explanation:

Problem 6

Step1: Use angle - addition postulate

Step2: Simplify the left - hand side of the equation

Step3: Solve for $x$

Step4: Find the measure of each angle

Answer:

Problem 7