Question

1. if m∠lmp is 11 degrees more than m∠nmp and m∠nml = 137°, find each measure.
2. if m∠abc is one degree less than three times m∠abd and m∠dbc = 47°, find each measure.
3. if (overline{qs}) bisects ∠pqt, m∠sqt=(8x - 25)°, m∠pqt=(9x + 34)°, and m∠sqr = 112°, find each measure.
4. 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: Set up equation 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 $x+(x + 11)=137$.
Combining like - terms gives $2x+11 = 137$.
Subtract 11 from both sides: $2x=137 - 11=126$.
Divide both sides by 2: $x=\frac{126}{2}=63$.
So $m\angle NMP = 63^{\circ}$ and $m\angle LMP=63 + 11=74^{\circ}$.

Step2: Set up equation 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: $3x-x-1=x-x + 47$, 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 $m\angle ABC=3\times24-1=71^{\circ}$.

Step3: Set up equation for problem 9

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

Step4: Set up equation for problem 10

Since $\overline{DE}$ bisects $\angle GDH$, $m\angle GDE=m\angle EDH$. Given $m\angle GDE=(8x - 1)^{\circ}$ and $m\angle EDH=(6x + 15)^{\circ}$, we have the equation $8x-1=6x + 15$.
Subtract $6x$ from both sides: $8x-6x-1=6x-6x + 15$, which gives $2x-1 = 15$.
Add 1 to both sides: $2x=15 + 1=16$.
Divide both sides by 2: $x = 8$.
$m\angle GDH=2m\angle GDE=2(8\times8 - 1)=2\times63 = 126^{\circ}$.
$m\angle FDH=180-(m\angle CDF + m\angle GDE)=180-(43+(8\times8 - 1))=180-(43 + 63)=74^{\circ}$.
$m\angle FDE=m\angle FDH+m\angle EDH=74+(6\times8 + 15)=74 + 63=137^{\circ}$.

Answer:

1. $m\angle LMP = 74^{\circ}$, $m\angle NMP = 63^{\circ}$
2. $m\angle ABD = 24^{\circ}$, $m\angle ABC = 71^{\circ}$
3. $x = 12$, $m\angle PQS = 71^{\circ}$, $m\angle PQT = 142^{\circ}$, $m\angle TQR = 41^{\circ}$
4. $x = 8$, $m\angle GDH = 126^{\circ}$, $m\angle FDH = 74^{\circ}$, $m\angle FDE = 137^{\circ}$