Question

points p, q, and s are collinear. if a ray qt bisects ∠rqs, what is m∠tqs? (3x - 5)° (x + 1)° m∠tqs =

Explanation:

Step1: Use angle - sum property of collinear points

Since points $P$, $Q$, and $S$ are collinear, $\angle PQR+\angle RQS = 180^{\circ}$. So, $(3x - 5)+(x + 1)=180$.
Combining like - terms: $3x+x-5 + 1=180$, which simplifies to $4x-4 = 180$.
Adding 4 to both sides: $4x=180 + 4=184$.
Dividing both sides by 4: $x=\frac{184}{4}=46$.

Step2: Find $\angle RQS$

Substitute $x = 46$ into the expression for $\angle RQS=(x + 1)^{\circ}$.
$\angle RQS=(46 + 1)^{\circ}=47^{\circ}$.

Step3: Use the angle - bisector property

Since ray $QT$ bisects $\angle RQS$, then $m\angle TQS=\frac{1}{2}m\angle RQS$.
$m\angle TQS=\frac{1}{2}\times47^{\circ}=23.5^{\circ}$.

Answer:

$23.5^{\circ}$