Question

1. if qs bisects ∠pqt, m∠sqt=(8x - 25)°, m∠pqt=(9x + 34)°, and m∠sqr = 112°, find each measure.

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: Solve the equation for $x$

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: $7x-50 + 50=34 + 50$, so $7x=84$.
Divide both sides by 7: $x = 12$.

Step3: Find $m\angle PQS$

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

Step4: Find $m\angle PQT$

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

Step5: 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:

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