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: For problem 6, use angle - addition postulate

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

Step2: For problem 7, let $m\angle NMP=x$. Then $m\angle LMP=x + 11$.

Since $m\angle NML=m\angle NMP+m\angle LMP$ and $m\angle NML = 137^{\circ}$, we have $x+(x + 11)=137$.
Combine like terms: $2x+11 = 137$.
Subtract 11 from both sides: $2x+11-11=137-11$, so $2x=126$.
Divide both sides by 2: $x = 63$.
$m\angle LMP=x + 11=63+11=74^{\circ}$.
$m\angle NMP=63^{\circ}$.

Step3: 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 $3x-1=x + 47$.
Subtract $x$ from both sides: $3x-x-1=x-x + 47$, so $2x-1 = 47$.
Add 1 to both sides: $2x-1+1=47+1$, so $2x=48$.
Divide both sides by 2: $x = 24$.
$m\angle ABD=24^{\circ}$.
$m\angle ABC=3x-1=3\times24-1=72 - 1=71^{\circ}$.

Step4: For problem 9, since $\overrightarrow{QS}$ bisects $\angle PQT$, then $m\angle PQS=m\angle SQT$ and $m\angle PQT = 2m\angle SQT$.

We know $m\angle SQT=(8x - 25)^{\circ}$ and $m\angle PQT=(9x + 34)^{\circ}$, so $9x + 34=2(8x - 25)$.
Expand the right - hand side: $9x + 34=16x-50$.
Subtract $9x$ from both sides: $9x-9x + 34=16x-9x-50$, so $34=7x-50$.
Add 50 to both sides: $34 + 50=7x-50+50$, so $84=7x$.
Divide both sides by 7: $x = 12$.
$m\angle PQS=8x-25=8\times12-25=96-25=71^{\circ}$.
$m\angle PQT=9x + 34=9\times12+34=108+34=142^{\circ}$.
$m\angle TQR=112^{\circ}-71^{\circ}=41^{\circ}$.

Step5: For problem 10, since $\overrightarrow{DE}$ bisects $\angle GDH$, then $m\angle GDE=m\angle EDH$.

So, $8x-1=6x + 15$.
Subtract $6x$ from both sides: $8x-6x-1=6x-6x + 15$, so $2x-1 = 15$.
Add 1 to both sides: $2x-1+1=15+1$, so $2x=16$.
Divide both sides by 2: $x = 8$.
$m\angle GDH=2m\angle GDE=2(8x-1)=2(8\times8-1)=2(64 - 1)=126^{\circ}$.
$m\angle FDE=180^{\circ}-43^{\circ}-m\angle GDH/2=180 - 43-63=74^{\circ}$.

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