Question

1. if m∠abc is one degree less than three times m∠abd and m∠dbc = 47°, find each measure. m∠abd = m∠abc = 9. if (overline{qs}) bisects ∠pqt, m∠sqt=(8x - 25)°, m∠pqt=(9x + 34)°, and m∠sqr = 112°, find each measure. x = m∠pqs = m∠pqt = m∠tqr =

Explanation:

Step1: Use angle - bisector property

Since $\overline{QS}$ 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 $16x-50 = 9x + 34$. Then, 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 $m\angle PQT$

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

Step4: Find $m\angle PQS$

Since $\overline{QS}$ bisects $\angle PQT$, $m\angle PQS=m\angle SQT$. Substitute $x = 12$ into the expression for $m\angle SQT$: $m\angle SQT=(8x - 25)^{\circ}=(8\times12-25)^{\circ}=(96 - 25)^{\circ}=71^{\circ}$, so $m\angle PQS = 71^{\circ}$.

Step5: Find $m\angle TQR$

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

Answer:

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