Question

1. if (overline{qs}) bisects (angle pqt), (mangle sqt=(8x - 25)^{circ}), (mangle pqt=(9x + 34)^{circ}), and (mangle sqr = 112^{circ}), find each measure.(x=) (mangle pqs=) (mangle pqt=) (mangle tqr=)10. if (angle cde) is a straight angle, (overline{de}) bisects (angle gdh), (mangle gde=(8x - 1)^{circ}), (mangle edh=(6x + 15)^{circ}), (mangle cdf = 43^{circ}), find each measure.(x=) (mangle gdh=) (mangle fdh=) (mangle fde=)

Explanation:

Step1: Set up equation for problem 9

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

Step2: Solve for $x$ in problem 9

Subtract $8x$ from both sides: $- 25=x + 34$. Then subtract 34 from both sides: $x=-59$.

Step3: Find angle - measures in problem 9

$m\angle PQS=9x + 34=9(-59)+34=-531 + 34=-497$ (This is incorrect, there is a sign - error in the setup. It should be $9x+34 + 8x - 25=112$).
Correcting:
$17x+9 = 112$, $17x=112 - 9=103$, $x=\frac{103}{17}\approx6.06$.
$m\angle PQS=9x + 34=9\times\frac{103}{17}+34=\frac{927}{17}+34=\frac{927+578}{17}=\frac{1505}{17}\approx88.53^{\circ}$
$m\angle SQT=8x - 25=8\times\frac{103}{17}-25=\frac{824}{17}-25=\frac{824 - 425}{17}=\frac{399}{17}\approx23.47^{\circ}$
$m\angle TQR = 112-(8x - 25)=112-(8\times\frac{103}{17}-25)=112-\frac{824}{17}+25=\frac{1904 - 824+425}{17}=\frac{1505}{17}\approx88.53^{\circ}$

Step4: Set up equation for problem 10

Since $\angle CDE$ is a straight - angle ($180^{\circ}$) and $\overline{DE}$ bisects $\angle GDH$, and $m\angle CDF = 43^{\circ}$.
Let $m\angle GDE=(8x - 1)^{\circ}$ and $m\angle EDH=(6x + 15)^{\circ}$. Then $m\angle GDE=m\angle EDH$. So, $8x-1 = 6x + 15$.

Step5: Solve for $x$ in problem 10

Subtract $6x$ from both sides: $2x-1 = 15$. Add 1 to both sides: $2x=16$, $x = 8$.

Step6: Find angle - measures in problem 10

$m\angle GDH=2m\angle GDE$ (because $DE$ bisects $\angle GDH$).
$m\angle GDE=8x - 1=8\times8 - 1=63^{\circ}$
$m\angle GDH = 126^{\circ}$
$m\angle FDH=180 - 43-63=74^{\circ}$
$m\angle FDE=63^{\circ}$

Answer:

For problem 9:
$x=\frac{103}{17}$, $m\angle PQS=\frac{1505}{17}^{\circ}$, $m\angle SQT=\frac{399}{17}^{\circ}$, $m\angle TQR=\frac{1505}{17}^{\circ}$
For problem 10:
$x = 8$, $m\angle GDH = 126^{\circ}$, $m\angle FDH=74^{\circ}$, $m\angle FDE=63^{\circ}$