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.
2. 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: Use angle - bisector property for problem 9

Since $\overline{QS}$ bisects $\angle PQT$, then $m\angle SQT=\frac{1}{2}m\angle PQT$. So, $8x - 25=\frac{1}{2}(9x + 34)$.
\[

$$\begin{align*} 8x-25&=\frac{9x + 34}{2}\\ 2(8x - 25)&=9x+34\\ 16x-50&=9x + 34\\ 16x-9x&=34 + 50\\ 7x&=84\\ x&=12 \end{align*}$$

\]

Step2: Find $m\angle PQS$

Since $\overline{QS}$ bisects $\angle PQT$ and $m\angle SQT = 8x-25$, when $x = 12$, $m\angle SQT=8\times12-25=96 - 25=71^{\circ}$, and $m\angle PQS=m\angle SQT = 71^{\circ}$

Step3: Find $m\angle PQT$

$m\angle PQT=9x + 34$, when $x = 12$, $m\angle PQT=9\times12+34=108 + 34=142^{\circ}$

Step4: Find $m\angle TQR$

$m\angle TQR=m\angle SQR - m\angle SQT$, $m\angle SQR = 112^{\circ}$, $m\angle SQT=71^{\circ}$, so $m\angle TQR=112^{\circ}-71^{\circ}=41^{\circ}$

Step5: Use angle - bisector property for problem 10

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

$$\begin{align*} 8x-6x&=15 + 1\\ 2x&=16\\ x&=8 \end{align*}$$

\]

Step6: Find $m\angle GDH$

$m\angle GDH=m\angle GDE+m\angle EDH$, $m\angle GDE=8x - 1$, $m\angle EDH=6x + 15$. When $x = 8$, $m\angle GDE=8\times8-1=63^{\circ}$, $m\angle EDH=6\times8 + 15=63^{\circ}$, so $m\angle GDH=63^{\circ}+63^{\circ}=126^{\circ}$

Step7: Find $m\angle FDH$

$\angle CDE$ is a straight - angle ($180^{\circ}$), $m\angle CDF = 43^{\circ}$, $m\angle FDH=180^{\circ}-m\angle CDF - m\angle EDH$. Since $m\angle EDH = 63^{\circ}$ and $m\angle CDF=43^{\circ}$, then $m\angle FDH=180^{\circ}-43^{\circ}-63^{\circ}=74^{\circ}$

Step8: Find $m\angle FDE$

$m\angle FDE=m\angle FDH+m\angle EDH$, $m\angle FDH = 74^{\circ}$, $m\angle EDH=63^{\circ}$, so $m\angle FDE=74^{\circ}+63^{\circ}=137^{\circ}$

Answer:

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

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