Question

use the protractor below to find the requested angle measure. assume all points are located on the 5s or 10s (ex: 20° not 18°). then, classify the angle.

1. m∠agc =
2. m∠cge =
3. m∠agb + m∠dge =
4. m∠cgf - m∠bga =
5. m∠bgf - m∠dge =

refer to the figure at the right for # 26 - 29

1. how many planes are shown in the figure?
2. name four points on plane 𝒜
3. name three collinear points.
4. name three non - collinear points.

for 30 - 31, use the figure to the right. qs bisects ∠pqt, (overrightarrow{qp}) and (overrightarrow{qr}) are opposite rays.

1. if m∠pqs = 5x - 11 and m∠sqt = 2x + 22,

a. find x
b. find m∠pqt

1. if m∠sqt = 2x + 5 and m∠tqr = 112°

a. find x.
b. find m∠tqp

Explanation:

Step1: Measure $\angle AGC$ with protractor

$m\angle AGC = 130^{\circ}$

Step2: Measure $\angle CGE$ with protractor

$m\angle CGE = 50^{\circ}$

Step3: Measure $\angle AGB$ and $\angle DGE$ with protractor

$m\angle AGB = 40^{\circ}$, $m\angle DGE = 40^{\circ}$, so $m\angle AGB + m\angle DGE=40^{\circ}+ 40^{\circ}=80^{\circ}$

Step4: Measure $\angle CGF$ and $\angle BGA$ with protractor

$m\angle CGF = 140^{\circ}$, $m\angle BGA = 40^{\circ}$, so $m\angle CGF - m\angle BGA=140^{\circ}-40^{\circ}=100^{\circ}$

Step5: Measure $\angle BGF$ and $\angle DGE$ with protractor

$m\angle BGF = 180^{\circ}$, $m\angle DGE = 40^{\circ}$, so $m\angle BGF - m\angle DGE=180^{\circ}-40^{\circ}=140^{\circ}$

Step6: Count planes in second - figure

There are 2 planes shown (plane $\mathcal{A}$ and the plane of the top - face of the solid)

Step7: Name points on plane $\mathcal{A}$

Four points on plane $\mathcal{A}$ are $M$, $N$, $R$, $X$ (answers may vary)

Step8: Name collinear points

Three collinear points are $W$, $S$, $M$ (answers may vary)

Step9: Name non - collinear points

Three non - collinear points are $M$, $N$, $P$ (answers may vary)

Step10: Solve for $x$ in $\angle PQS$ and $\angle SQT$ problem

Since $QS$ bisects $\angle PQT$, $m\angle PQS=m\angle SQT$. So $5x - 11=2x + 22$.
Subtract $2x$ from both sides: $5x-2x - 11=2x-2x + 22$, $3x-11 = 22$.
Add 11 to both sides: $3x-11 + 11=22 + 11$, $3x=33$.
Divide both sides by 3: $x = 11$.

Step11: Find $m\angle PQT$

$m\angle PQT=m\angle PQS + m\angle SQT$. Since $x = 11$, $m\angle PQS=5x-11=5\times11-11 = 44$, $m\angle SQT=2x + 22=2\times11+22 = 44$, so $m\angle PQT=44^{\circ}+44^{\circ}=88^{\circ}$

Step12: Solve for $x$ in $\angle SQT$ and $\angle TQR$ problem

Since $\angle SQT+\angle TQR = 180^{\circ}$ (linear pair), $2x + 5+112=180$.
$2x+117 = 180$.
Subtract 117 from both sides: $2x=180 - 117=63$.
Divide both sides by 2: $x = 31.5$

Step13: Find $m\angle TQP$

There is a typo in the problem, it should be $m\angle TQP$. Since $\angle SQT = 2x+5$ and $x = 31.5$, $m\angle SQT=2\times31.5+5=63 + 5=68^{\circ}$. And since $\angle PQT$ and $\angle TQR$ are a linear pair and $\angle TQR = 112^{\circ}$, $m\angle TQP = 68^{\circ}$

Answer:

1. $130^{\circ}$
2. $50^{\circ}$
3. $80^{\circ}$
4. $100^{\circ}$
5. $140^{\circ}$
6. 2
7. $M$, $N$, $R$, $X$ (answers may vary)
8. $W$, $S$, $M$ (answers may vary)
9. $M$, $N$, $P$ (answers may vary)
10. a. $x = 11$

b. $88^{\circ}$

1. a. $x = 31.5$

b. $68^{\circ}$