Given that $QS$ bisects $\angle PQT$, let's break down the problem:

• $\angle PQT = 9x + 34$
• Since $QS$ bisects $\angle PQT$, the two angles formed, $\angle PQS$ and $\angle SQT$, are equal.
• So, $\angle PQS = \angle SQT = \frac{1}{2} \times \angle PQT$.
• $\angle SQT = 8x - 25$ (given)
• $\angle SQR = 112^\circ$ (given)

Let's find $x$ first by setting $\angle SQT = \frac{1}{2} \times \angle PQT$:

$8x - 25 = \frac{1}{2} \times (9x + 34)$

Now, solve for $x$:

$8x - 25 = \frac{9x + 34}{2}$

Multiply both sides by 2 to eliminate the fraction:

$16x - 50 = 9x + 34$

Now, solve for $x$:

$16x - 9x = 34 + 50$ $7x = 84$ $x = 12$

Now that we have $x = 12$, we can find the measures of the angles:

1. $\angle PQT$: $\angle PQT = 9x + 34 = 9(12) + 34 = 108 + 34 = 142^\circ$

2. $\angle SQT$ (which is also $\angle PQS$ because $QS$ bisects $\angle PQT$): $\angle SQT = 8x - 25 = 8(12) - 25 = 96 - 25 = 71^\circ$

3. $\angle SQR$ is given as $112^\circ$.

Thus, the angles are:

• $\angle PQT = 142^\circ$
• $\angle SQT = \angle PQS = 71^\circ$
• $\angle SQR = 112^\circ$ (already given)

Would you like any further clarification on these steps?

Here are 5 related questions:

1. What if $\angle SQT$ was given instead as $7x - 15$? How would the angles change?
2. How can you use the sum of angles in a triangle to verify the calculation?
3. If $\angle SQR$ was changed, how would that affect the overall solution?
4. What if $QS$ didn't bisect $\angle PQT$, but instead was given as another angle?
5. Can you solve a similar problem if an exterior angle and one interior angle are given?

Tip: When working with angle bisectors, always remember that the two resulting angles are equal, which often leads to setting up a helpful equation.