22. the measure of $\angle g$ is six more than twice the measure of $\angle h$. if $\angle g$ and $\angle h$ are complementary angles, find $m\angle h$.
23. the measure of $\angle z$ is twelve less than five times the measure of $\angle y$. if $\angle y$ and $\angle z$ form a linear pair, find $m\angle z$.
24. if $\overrightarrow{bd}$ bisects $\angle abc$, $m\angle dbc = 79^\circ$, and $m\angle abc = 9x - 4$, find the value of $x$.
25. if $\overrightarrow{qs}$ bisects $\angle pqr$, $m\angle pqs = 7x - 6$, and $m\angle sqr = 4x + 15$, find $m\angle pqt$.

Question

Accepted Answer

### Problem 22
# Explanation:
## Step1: Define variables and relationship
Let $ m\angle H = x $. Then $ m\angle G = 2x + 6 $. Since they are complementary, $ m\angle G + m\angle H = 90^\circ $.
$$ (2x + 6) + x = 90 $$
## Step2: Solve for x
Combine like terms:
$$ 3x + 6 = 90 $$
Subtract 6 from both sides:
$$ 3x = 84 $$
Divide by 3:
$$ x = 28 $$
# Answer:
$ 28^\circ $

### Problem 23
# Explanation:
## Step1: Define variables and relationship
Let $ m\angle Y = x $. Then $ m\angle Z = 5x - 12 $. A linear pair sums to $ 180^\circ $, so $ x + (5x - 12) = 180 $.
$$ 6x - 12 = 180 $$
## Step2: Solve for x
Add 12 to both sides:
$$ 6x = 192 $$
Divide by 6:
$$ x = 32 $$
## Step3: Find $ m\angle Z $
Substitute $ x = 32 $ into $ 5x - 12 $:
$$ 5(32) - 12 = 160 - 12 = 148 $$
# Answer:
$ 148^\circ $

### Problem 24
# Explanation:
## Step1: Use angle bisector property
Since $ \overrightarrow{BD} $ bisects $ \angle ABC $, $ m\angle ABC = 2 \times m\angle DBC $. So $ 9x - 4 = 2 \times 79 $.
$$ 9x - 4 = 158 $$
## Step2: Solve for x
Add 4 to both sides:
$$ 9x = 162 $$
Divide by 9:
$$ x = 18 $$
# Answer:
$ 18 $

### Problem 25
# Explanation:
## Step1: Use angle bisector property
Since $ \overrightarrow{QS} $ bisects $ \angle PQR $, $ m\angle PQS = m\angle SQR $. So $ 7x - 6 = 4x + 15 $.
$$ 7x - 4x = 15 + 6 $$
$$ 3x = 21 $$
## Step2: Solve for x
Divide by 3:
$$ x = 7 $$
## Step3: Find $ m\angle PQS $
Substitute $ x = 7 $ into $ 7x - 6 $:
$$ 7(7) - 6 = 49 - 6 = 43^\circ $$
## Step4: Find $ m\angle PQT $
$ \angle PQT $ is vertical to $ \angle RQW $, but first, $ \angle PQR = 2 \times 43 = 86^\circ $. Since $ \angle PQT $ and $ \angle RQW $ are vertical, and $ \angle PQR + \angle RQW = 180^\circ $? Wait, no. Wait, $ \overrightarrow{QT} $ and $ \overrightarrow{QW} $ are opposite? Wait, actually, since $ \angle PQS = 43^\circ $, $ \angle SQR = 43^\circ $. Then $ \angle PQT $ is supplementary to $ \angle PQR $? Wait, no. Wait, looking at the diagram, $ \angle PQT $ and $ \angle RQW $ are vertical angles, and $ \angle PQR $ and $ \angle TQW $ are vertical angles. Wait, actually, since $ \overrightarrow{QS} $ bisects $ \angle PQR $, $ m\angle PQR = 2 \times 43 = 86^\circ $. Then $ \angle PQT $ is $ 180^\circ - 86^\circ = 94^\circ $? Wait, no. Wait, maybe $ \angle PQT $ is equal to $ \angle RQW $, and $ \angle PQR + \angle RQW = 180^\circ $? Wait, no, let's re-express.

Wait, when $ \overrightarrow{QS} $ bisects $ \angle PQR $, $ m\angle PQS = m\angle SQR = 43^\circ $, so $ m\angle PQR = 86^\circ $. Then, since $ \angle PQT $ and $ \angle RQW $ are vertical angles, and $ \angle PQR $ and $ \angle TQW $ are vertical angles. Wait, actually, the lines $ PR $ and $ TW $ intersect at $ Q $, so $ \angle PQT $ and $ \angle RQW $ are vertical, and $ \angle PQR $ and $ \angle TQW $ are vertical. So $ \angle PQT + \angle PQR = 180^\circ $? No, that's not right. Wait, maybe $ \angle PQT $ is supplementary to $ \angle PQR $? Wait, no, let's check the angles.

Wait, the correct approach: since $ \overrightarrow{QS} $ bisects $ \angle PQR $, $ m\angle PQS = m\angle SQR = 43^\circ $, so $ m\angle PQR = 86^\circ $. Then, $ \angle PQT $ is equal to $ 180^\circ - 86^\circ = 94^\circ $? Wait, no, that's if they are supplementary. Wait, actually, looking at the diagram, $ \angle PQT $ and $ \angle RQW $ are vertical angles, and $ \angle PQR $ and $ \angle TQW $ are vertical angles. So $ \angle PQT = \angle RQW $, and $ \angle PQR + \angle RQW = 180^\circ $, so $ \angle PQT = 180^\circ - 86^\circ = 94^\circ $. Wait, but let's confirm.

Wait, another way: since $ \overrightarrow{QS} $ bisects $ \angle PQR $, $ m\angle PQS = m\angle SQR = 43^\circ $, so $ m\angle PQR = 86^\circ $. Then, $ \angle PQT $ is adjacent to $ \angle PQR $ forming a linear pair? Wait, no, the diagram shows $ P, S, R $ on one line, and $ T, Q, W $ on another? Wait, no, the diagram has $ P, S, R $ with $ S $ between $ P $ and $ R $, and $ T, Q, W $ with $ Q $ as the vertex. Wait, actually, $ \overrightarrow{QS} $ bisects $ \angle PQR $, so $ \angle PQS = \angle SQR = 43^\circ $. Then, $ \angle PQT $ is vertical to $ \angle RQW $, and $ \angle PQR + \angle RQW = 180^\circ $, so $ \angle RQW = 180^\circ - 86^\circ = 94^\circ $, so $ \angle PQT = 94^\circ $.

Wait, but let's do it step by step.

First, solve for $ x $: $ 7x - 6 = 4x + 15 $ gives $ 3x = 21 $, so $ x = 7 $. Then $ m\angle PQS = 7(7) - 6 = 43^\circ $, so $ m\angle SQR = 43^\circ $, so $ m\angle PQR = 43 + 43 = 86^\circ $. Now, $ \angle PQT $ and $ \angle RQW $ are vertical angles, and $ \angle PQR $ and $ \angle TQW $ are vertical angles. Also, $ \angle PQT + \angle PQR = 180^\circ $ (since they form a linear pair? Wait, no, if $ P, Q, R $ are on a line? No, $ \overrightarrow{QS} $ is a bisector, so $ P, S, R $ are on a line? Wait, the diagram shows $ S $ between $ P $ and $ R $, so $ PR $ is a straight line, so $ \angle PQS + \angle SQR = 180^\circ $? No, that can't be, because they are angles at $ Q $. Wait, I think I made a mistake.

Wait, no, the problem says $ \overrightarrow{QS} $ bisects $ \angle PQR $, so $ \angle PQS = \angle SQR $, and $ \angle PQR = \angle PQS + \angle SQR $. Then, $ \angle PQT $ is vertical to $ \angle RQW $, and $ \angle PQR $ is vertical to $ \angle TQW $. Also, $ \angle PQT + \angle TQW = 180^\circ $ (since they are adjacent and form a linear pair). Wait, no, $ \angle PQT $ and $ \angle RQW $ are vertical, so they are equal. $ \angle PQR $ and $ \angle TQW $ are vertical, so they are equal. Then, since $ PR $ and $ TW $ intersect at $ Q $, the sum of adjacent angles is $ 180^\circ $. So $ \angle PQT + \angle PQR = 180^\circ $? No, that would be if $ T, Q, P $ are on a line, but the diagram shows $ T $ and $ W $ as opposite directions. Wait, maybe the correct approach is that $ \angle PQT $ is supplementary to $ \angle PQR $. Wait, let's check the angles again.

Wait, when $ x = 7 $, $ m\angle PQS = 43^\circ $, $ m\angle SQR = 43^\circ $, so $ m\angle PQR = 86^\circ $. Then, $ \angle PQT $ is equal to $ 180^\circ - 86^\circ = 94^\circ $, because $ \angle PQT $ and $ \angle PQR $ form a linear pair? Wait, no, that would be if $ T, Q, R $ are on a line, but the diagram shows $ T $ and $ W $ as opposite. Wait, maybe the diagram is such that $ PR $ and $ TW $ are intersecting lines, so $ \angle PQT $ and $ \angle RQW $ are vertical, and $ \angle PQR $ and $ \angle TQW $ are vertical. Then, $ \angle PQT + \angle PQR = 180^\circ $ because they are adjacent angles on a straight line. Wait, yes, if $ T, Q, W $ is a straight line, and $ P, Q, R $ is a straight line, then they intersect at $ Q $, forming vertical angles. So $ \angle PQT $ and $ \angle RQW $ are vertical, $ \angle PQR $ and $ \angle TQW $ are vertical. Then, $ \angle PQT + \angle PQR = 180^\circ $ (since $ T, Q, W $ is a straight line, so $ \angle PQT + \angle TQP = 180^\circ $? No, I'm getting confused.

Wait, let's start over. The key is that $ \overrightarrow{QS} $ bisects $ \angle PQR $, so $ m\angle PQS = m\angle SQR $. We found $ x = 7 $, so $ m\angle PQS = 43^\circ $, $ m\angle SQR = 43^\circ $, so $ m\angle PQR = 86^\circ $. Now, $ \angle PQT $ is vertical to $ \angle RQW $, and $ \angle PQR $ is vertical to $ \angle TQW $. Also, $ \angle PQT + \angle TQW = 180^\circ $ (since they are adjacent and form a linear pair). But $ \angle TQW = \angle PQR = 86^\circ $, so $ \angle PQT = 180^\circ - 86^\circ = 94^\circ $. Yes, that makes sense. So $ m\angle PQT = 94^\circ $.

# Answer:
$ 94^\circ $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 22

Step1: Define variables and relationship

Step2: Solve for x

Step1: Define variables and relationship

Step2: Solve for x

Step3: Find \( m\angle Z \)

Step1: Use angle bisector property

Step2: Solve for x

Answer:

Problem 23