Question

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

Explanation:

Step1: Use angle - addition property for problem 6

Since $\angle CDF+\angle FDE=\angle CDE$, we have the equation $(3x + 14)+(5x - 2)=10x-18$.
First, simplify the left - hand side:
$3x+14 + 5x-2=8x + 12$.
So the equation becomes $8x + 12=10x-18$.
Subtract $8x$ from both sides:
$12=10x-8x-18$.
$12 = 2x-18$.
Add 18 to both sides:
$12 + 18=2x$.
$30=2x$.
Divide both sides by 2:
$x = 15$.
Then find the angle measures:
$m\angle CDF=3x + 14=3\times15+14=45 + 14=59^{\circ}$.
$m\angle FDE=5x - 2=5\times15-2=75 - 2=73^{\circ}$.
$m\angle CDE=10x-18=10\times15-18=150 - 18=132^{\circ}$.

Step2: Use angle - addition property for problem 7

Let $m\angle NMP=x$. Then $m\angle LMP=x + 11$.
Since $m\angle NML=m\angle LMP+m\angle NMP$ and $m\angle NML = 137^{\circ}$, we have the equation $x+(x + 11)=137$.
Simplify the left - hand side: $2x+11 = 137$.
Subtract 11 from both sides: $2x=137 - 11=126$.
Divide both sides by 2: $x = 63$.
So $m\angle NMP=63^{\circ}$ and $m\angle LMP=63 + 11=74^{\circ}$.

Step3: Use angle - addition property for problem 8

Let $m\angle ABD=x$. Then $m\angle ABC=3x-1$.
Since $m\angle ABC=m\angle ABD+m\angle DBC$ and $m\angle DBC = 47^{\circ}$, we have the equation $3x-1=x + 47$.
Subtract $x$ from both sides: $3x-x-1=47$.
$2x-1=47$.
Add 1 to both sides: $2x=47 + 1=48$.
Divide both sides by 2: $x = 24$.
So $m\angle ABD=24^{\circ}$ and $m\angle ABC=3\times24-1=72 - 1=71^{\circ}$.

Step4: Use angle - bisector property for problem 9

Since $\overline{QS}$ bisects $\angle PQT$, then $m\angle SQT=\frac{1}{2}m\angle PQT$.
We have the equation $8x-25=\frac{1}{2}(9x + 34)$.
Multiply both sides by 2 to get rid of the fraction: $2(8x-25)=9x + 34$.
$16x-50=9x + 34$.
Subtract $9x$ from both sides: $16x-9x-50=34$.
$7x-50=34$.
Add 50 to both sides: $7x=34 + 50=84$.
Divide both sides by 7: $x = 12$.
$m\angle PQS=m\angle SQT=8x-25=8\times12-25=96 - 25=71^{\circ}$.
$m\angle PQT=9x + 34=9\times12+34=108+34=142^{\circ}$.
$m\angle TQR=m\angle SQR - m\angle SQT=112-71 = 41^{\circ}$.

Step5: Use angle - bisector property for problem 10

Since $\overline{DE}$ bisects $\angle GDH$, then $m\angle GDE=m\angle EDH$.
So we have the equation $8x-1=6x + 15$.
Subtract $6x$ from both sides: $8x-6x-1=15$.
$2x-1=15$.
Add 1 to both sides: $2x=15 + 1=16$.
Divide both sides by 2: $x = 8$.
$m\angle GDH=2m\angle GDE=2(8x-1)=2(8\times8-1)=2(64 - 1)=126^{\circ}$.
$m\angle FDH=m\angle CDE - m\angle CDF - m\angle EDH$. Since $\angle CDE$ is a straight angle ($m\angle CDE = 180^{\circ}$), $m\angle CDF = 43^{\circ}$, and $m\angle EDH=6x + 15=6\times8+15=48 + 15=63^{\circ}$.
$m\angle FDH=180-43 - 63=74^{\circ}$.
$m\angle FDE=m\angle CDE - m\angle CDF=180-43=137^{\circ}$.

Answer:

Problem 6:
$x = 15$
$m\angle CDF=59^{\circ}$
$m\angle FDE=73^{\circ}$
$m\angle CDE=132^{\circ}$

Problem 7:
$m\angle LMP=74^{\circ}$
$m\angle NMP=63^{\circ}$

Problem 8:
$m\angle ABD=24^{\circ}$
$m\angle ABC=71^{\circ}$

Problem 9:
$x = 12$
$m\angle PQS=71^{\circ}$
$m\angle PQT=142^{\circ}$
$m\angle TQR=41^{\circ}$

Problem 10:
$x = 8$
$m\angle GDH=126^{\circ}$
$m\angle FDH=74^{\circ}$
$m\angle FDE=137^{\circ}$