Question

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

Explanation:

Step1: Set up equation for problem 6

Since $\angle CDF+\angle FDE=\angle CDE$, we have $(3x + 14)+(5x - 2)=10x-18$.

Step2: Simplify left - hand side

Combine 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 $12 = 2x-18$. Add 18 to both sides: $12 + 18=2x$, so $30 = 2x$. Divide by 2, $x = 15$.

Step4: Find angle measures

$m\angle CDF=3x + 14=3\times15+14=45 + 14=59^{\circ}$.
$m\angle FDE=5x - 2=5\times15-2=75 - 2=73^{\circ}$.
$m\angle CDE=10x-18=10\times15-18=150 - 18=132^{\circ}$.

Step5: Set up equation for problem 7

Let $m\angle NMP=y$, then $m\angle LMP=y + 11$. Since $m\angle NML=m\angle LMP+m\angle NMP$, we have $y+(y + 11)=137$.

Step6: Solve for $y$

Combine like terms: $2y+11 = 137$. Subtract 11 from both sides: $2y=137 - 11=126$. Divide by 2: $y = 63$. So $m\angle NMP=63^{\circ}$ and $m\angle LMP=63 + 11=74^{\circ}$.

Step7: Set up equation for problem 8

Let $m\angle ABD=z$, then $m\angle ABC=3z-1$. Since $m\angle ABC=m\angle ABD+\angle DBC$, we have $3z-1=z + 47$.

Step8: Solve for $z$

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

Step9: Set up equation for problem 9

Since $\overline{QS}$ bisects $\angle PQT$, $m\angle PQS=m\angle SQT$ and $m\angle PQT = 2m\angle SQT$. So $9x + 34=2(8x - 25)$.

Step10: Expand right - hand side

$9x+34=16x-50$.

Step11: Solve for $x$

Subtract $9x$ from both sides: $34=16x-9x-50$, so $34 = 7x-50$. Add 50 to both sides: $34 + 50=7x$, so $84 = 7x$, and $x = 12$.
$m\angle PQS=m\angle SQT=8x - 25=8\times12-25=96 - 25=71^{\circ}$.
$m\angle PQT=9x + 34=9\times12+34=108 + 34=142^{\circ}$.
$m\angle TQR=m\angle SQR - m\angle SQT=112-71 = 41^{\circ}$.

Step12: Set up equation for problem 10

Since $\overline{DE}$ bisects $\angle GDH$, $m\angle GDE=m\angle EDH$. So $8x-1=6x + 15$.

Step13: Solve for $x$

Subtract $6x$ from both sides: $8x-6x-1=15$, so $2x-1 = 15$. Add 1 to both sides: $2x=16$, divide by 2, $x = 8$.
$m\angle GDH=2m\angle GDE=2(8x - 1)=2(8\times8-1)=2(64 - 1)=126^{\circ}$.
$m\angle FDH=180 - m\angle CDF - m\angle GDE=180-43-(8\times8 - 1)=180-43 - 63=74^{\circ}$.
$m\angle FDE=m\angle FDH+m\angle HDE=74+(6\times8 + 15)=74+63=137^{\circ}$.

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