Question

1. if $overrightarrow{qs}$ bisects $angle pqt$, $mangle sqt=(8x - 25)^{circ}$, $mangle pqt=(9x + 34)^{circ}$, and $mangle sqr = 112^{circ}$, find each measure.

5.
x =
$mangle pqs=$
$mangle pqt=$
$mangle tqr=$

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{QS}$ bisects $\angle PQT$, we know that $m\angle SQT=\frac{1}{2}m\angle PQT$. So, $2(8x - 25)=9x + 34$.

Step2: Expand the left - hand side

Expand $2(8x - 25)$ to get $16x-50$. The equation becomes $16x-50 = 9x + 34$.

Step3: Solve for $x$

Subtract $9x$ from both sides: $16x-9x-50=9x - 9x+34$, which simplifies to $7x-50 = 34$. Then add 50 to both sides: $7x-50 + 50=34 + 50$, so $7x=84$. Divide both sides by 7: $x = 12$.

Step4: Find $m\angle PQS$

Since $m\angle PQS=m\angle SQT$ and $m\angle SQT=(8x - 25)^{\circ}$, substitute $x = 12$: $m\angle PQS=8\times12-25=96 - 25=71^{\circ}$.

Step5: Find $m\angle PQT$

Substitute $x = 12$ into $m\angle PQT=(9x + 34)^{\circ}$: $m\angle PQT=9\times12+34=108 + 34=142^{\circ}$.

Step6: Find $m\angle TQR$

We know that $m\angle SQR = 112^{\circ}$ and $m\angle SQT = 71^{\circ}$. Then $m\angle TQR=m\angle SQR - m\angle SQT=112-71 = 41^{\circ}$.

Answer:

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