Question

1. if m∠9 = 97° and m∠12 = 114°, find each measure.

a. m∠1 =
b. m∠2 =
c. m∠3 =
d. m∠4 =
e. m∠5 =
f. m∠6 =
g. m∠7 =
h. m∠8 =
i. m∠10 =
j. m∠11 =
k. m∠13 =
l. m∠14 =
m. m∠15 =
n. m∠16 = 3. if m∠2 = 98°, m∠3 = 23° and m∠8 = 70°, find each measure.
a. m∠1 =
d. m∠6 =
g. m∠10 =

Explanation:

Step1: Identify vertical - angle and supplementary - angle relationships

Vertical angles are equal. Supplementary angles add up to 180°.

Step2: For problem 2

a.

$\angle1$ and $\angle9$ are vertical angles. So $m\angle1 = m\angle9=97^{\circ}$.

b.

$\angle2$ and $\angle12$ are vertical angles. So $m\angle2 = m\angle12 = 114^{\circ}$.

c.

$\angle3$ and an angle supplementary to $\angle9$ and $\angle12$ combination. First, find the angle adjacent to $\angle9$ and $\angle12$. Let's assume the unknown adjacent angle to $\angle9$ and $\angle12$ is $x$. The sum of angles around a point is 360°. If we consider the intersection of the lines, we know that $m\angle3$: Since $\angle9 = 97^{\circ}$ and $\angle12=114^{\circ}$, and the sum of angles around the intersection is 360°, we can find the angle adjacent to $\angle9$ and $\angle12$ and then use vertical - angle or linear - pair relationships. But a simpler way is to note that if we consider the linear - pair and vertical - angle relationships, we know that $\angle3$ and an angle related to the non - overlapping part of $\angle9$ and $\angle12$. Since $\angle9 = 97^{\circ}$ and $\angle12 = 114^{\circ}$, and the sum of angles around a point is 360°, we can find that $m\angle3=149^{\circ}$.

d.

$\angle4$ and $\angle3$ are vertical angles. So $m\angle4 = m\angle3 = 149^{\circ}$.

e.

$\angle5$ and $\angle9$ are supplementary (linear - pair). So $m\angle5=180^{\circ}-97^{\circ}=83^{\circ}$.

f.

$\angle6$ and $\angle12$ are supplementary (linear - pair). So $m\angle6=180^{\circ}-114^{\circ}=66^{\circ}$.

g.

$\angle7$ and $\angle5$ are vertical angles. So $m\angle7 = m\angle5 = 83^{\circ}$.

h.

$\angle8$ and $\angle6$ are vertical angles. So $m\angle8 = m\angle6 = 66^{\circ}$.

i.

$\angle10$ and $\angle6$ are supplementary (linear - pair). So $m\angle10=180^{\circ}-66^{\circ}=114^{\circ}$.

j.

$\angle11$ and $\angle5$ are supplementary (linear - pair). So $m\angle11=180^{\circ}-83^{\circ}=97^{\circ}$.

k.

$\angle13$ and $\angle1$ are supplementary (linear - pair). So $m\angle13=180^{\circ}-97^{\circ}=83^{\circ}$.

l.

$\angle14$ and $\angle2$ are supplementary (linear - pair). So $m\angle14=180^{\circ}-114^{\circ}=66^{\circ}$.

m.

$\angle15$ and $\angle3$ are supplementary (linear - pair). So $m\angle15=180^{\circ}-149^{\circ}=31^{\circ}$.

n.

$\angle16$ and $\angle4$ are supplementary (linear - pair). So $m\angle16=180^{\circ}-149^{\circ}=31^{\circ}$.

Step3: For problem 3

a.

We know that the sum of angles in a triangle formed by $\angle1$, $\angle2$, and $\angle3$ (assuming they are related in a triangle - like formation by the intersection of lines) is 180°. But we can also use linear - pair and vertical - angle relationships. $\angle1$ and an angle supplementary to $\angle2$ and $\angle3$. First, find the angle adjacent to $\angle2$ and $\angle3$. Let $y$ be the angle adjacent to $\angle2$ and $\angle3$. We know that $m\angle1=180^{\circ}-(m\angle2 + m\angle3)$. Substituting $m\angle2 = 98^{\circ}$ and $m\angle3 = 23^{\circ}$, we get $m\angle1=180^{\circ}-(98^{\circ}+23^{\circ})=59^{\circ}$.

d.

$\angle6$ and $\angle2$ are vertical angles. So $m\angle6 = m\angle2 = 98^{\circ}$.

g.

$\angle10$ and $\angle3$ are vertical angles. So $m\angle10 = m\angle3 = 23^{\circ}$.

h.

$m\angle8 = 70^{\circ}$ (given).

Answer:

Problem 2

a. $m\angle1 = 97^{\circ}$
b. $m\angle2 = 114^{\circ}$
c. $m\angle3 = 149^{\circ}$
d. $m\angle4 = 149^{\circ}$
e. $m\angle5 = 83^{\circ}$
f. $m\angle6 = 66^{\circ}$
g. $m\angle7 = 83^{\circ}$
h. $m\angle8 = 66^{\circ}$
i. $m\angle10 = 114^{\circ}$
j. $m\angle11 = 97^{\circ}$
k. $m\angle13 = 83^{\circ}$
l. $m\angle14 = 66^{\circ}$
m. $m\angle15 = 31^{\circ}$
n. $m\angle16 = 31^{\circ}$

Problem 3

a. $m\angle1 = 59^{\circ}$
d. $m\angle6 = 98^{\circ}$
g. $m\angle10 = 23^{\circ}$
h. $m\angle8 = 70^{\circ}$