Question

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

Explanation:

Step1: Use angle - bisector property

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

Step2: Solve the equation for $x$

Multiply both sides of the equation $8x - 25=\frac{1}{2}(9x + 34)$ by 2 to get $2(8x - 25)=9x + 34$. Expand the left - hand side: $16x-50 = 9x + 34$. Subtract $9x$ from both sides: $16x-9x-50=9x - 9x+34$, which simplifies to $7x-50 = 34$. 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 the formula. $m\angle PQS=8\times12-25=96 - 25=71^{\circ}$.

Step4: Find $m\angle PQT$

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

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:

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