Question

1. if (mangle abc) is one degree less than three times (mangle abd) and (mangle dbc = 47^{circ}), find each measure.

(mangle abd=)
(mangle abc=)

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.

(x=)
(mangle pqs=)
(mangle pqt=)
(mangle tqr=)

Explanation:

Step1: Use angle - bisector property

Since $\overline{QS}$ bisects $\angle PQT$, we know that $m\angle SQT=\frac{1}{2}m\angle PQT$. 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 $7x-50 = 34$. Then 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 $m\angle PQT+m\angle TQR = 180^{\circ}$ (linear pair). Since $m\angle PQT = 142^{\circ}$, then $m\angle TQR=180^{\circ}-m\angle PQT=180 - 142=38^{\circ}$.

Answer:

$x = 12$, $m\angle PQT = 142^{\circ}$, $m\angle PQS = 71^{\circ}$, $m\angle TQR = 38^{\circ}$