Question

practice
example 1
use the figure to identify angles and parts of angles that satisfy each given condition.

1. name the vertex of ∠1.
2. name the sides of ∠4.
3. what is another name for ∠3?
4. what is another name for ∠cad?

example 2

1. in the figure, $overrightarrow{lf}$ and $overrightarrow{lk}$ are opposite rays. $overrightarrow{lg}$ bisects ∠flh. if $mangle flg=(14x + 5)^{circ}$ and $mangle hlg=(17x - 1)^{circ}$, find $mangle flh$.

in the figure, $overrightarrow{ba}$ and $overrightarrow{bc}$ are opposite rays. $overrightarrow{bh}$ bisects ∠ebc and $overrightarrow{be}$ bisects ∠abf.

1. if $mangle abe=(2n + 7)^{circ}$ and $mangle ebf=(4n - 13)^{circ}$, find $mangle abe$.
2. if $mangle ebh=(6x + 12)^{circ}$ and $mangle hbc=(8x - 10)^{circ}$, find $mangle ebh$.
3. if $mangle abf=(7b - 24)^{circ}$ and $mangle abe = 2b^{circ}$, find $mangle ebf$.
4. if $mangle ebc=(31a - 2)^{circ}$ and $mangle ebh=(4a + 45)^{circ}$, find $mangle hbc$.
5. if $mangle abf=(8w - 6)^{circ}$ and $mangle abe=2(w + 11)^{circ}$, find $mangle ebf$.
6. if $mangle ebc=(3r + 10)^{circ}$ and $mangle abe=(2r - 20)^{circ}$, find $mangle ebf$.

Explanation:

Step1: Recall angle - vertex definition

The vertex of an angle is the common endpoint of the rays that form the angle. For $\angle1$, the vertex is point $A$.

Step2: Recall angle - side definition

The sides of an angle are the two rays that form the angle. For $\angle4$, the sides are $\overrightarrow{CA}$ and $\overrightarrow{CD}$.

Step3: Find alternative angle name

$\angle3$ can also be named $\angle EDA$ since the vertex is $D$ and the rays are $\overrightarrow{DE}$ and $\overrightarrow{DA}$.

Step4: Find alternative angle name

$\angle CAD$ can also be named $\angle 4$ as it represents the same angle.

Step5: Use angle - bisector property

Since $\overrightarrow{LG}$ bisects $\angle FLH$, then $m\angle FLG=m\angle HLG$. So, $14x + 5=17x-1$.

Solve for $x$

$17x-14x=5 + 1$, $3x=6$, $x = 2$. Then $m\angle FLH=2m\angle FLG=2(14\times2 + 5)=2(28 + 5)=66^{\circ}$.

Step6: Use angle - bisector property

Since $\overrightarrow{BE}$ bisects $\angle ABF$, then $m\angle ABE=m\angle EBF$. So, $2n+7=4n - 13$.

Solve for $n$

$4n-2n=7 + 13$, $2n=20$, $n = 10$. Then $m\angle ABE=2\times10+7=27^{\circ}$.

Step7: Use angle - bisector property

Since $\overrightarrow{BH}$ bisects $\angle EBC$, then $m\angle EBH=m\angle HBC$. So, $6x + 12=8x-10$.

Solve for $x$

$8x-6x=12 + 10$, $2x=22$, $x = 11$. Then $m\angle EBH=6\times11+12=78^{\circ}$.

Step8: Since $m\angle ABF=m\angle ABE+m\angle EBF$

$7b-24=2b+m\angle EBF$. Also, since $\overrightarrow{BE}$ bisects $\angle ABF$ (from the given ray - bisecting relationship), $m\angle EBF=m\angle ABE$. So, $7b-24=2b+2b$, $7b-4b=24$, $3b=24$, $b = 8$. Then $m\angle EBF=2\times8=16^{\circ}$.

Step9: Since $m\angle EBC=2m\angle EBH$ (because $\overrightarrow{BH}$ bisects $\angle EBC$)

$31a-2=2(4a + 45)$.

Expand and solve for $a$

$31a-2=8a+90$, $31a-8a=90 + 2$, $23a=92$, $a = 4$. Then $m\angle EBH=4\times4+45=61^{\circ}$, and $m\angle HBC=m\angle EBH = 61^{\circ}$.

Step10: Since $m\angle ABF=m\angle ABE+m\angle EBF$

$8w-6=2(w + 11)+m\angle EBF$. Also, since $\overrightarrow{BE}$ bisects $\angle ABF$, $m\angle EBF=m\angle ABE$. So, $8w-6=2(w + 11)+2(w + 11)$.

Expand and solve for $w$

$8w-6=2w+22+2w + 22$, $8w-6=4w + 44$, $8w-4w=44 + 6$, $4w=50$, $w=\frac{25}{2}$. Then $m\angle EBF=2(\frac{25}{2}+11)=2\times\frac{25 + 22}{2}=47^{\circ}$.

Step11: Since $m\angle EBC+m\angle ABE = 180^{\circ}$ (linear - pair of angles)

$(3r + 10)+(2r-20)=180$.

Combine like terms and solve for $r$

$3r+2r+10-20=180$, $5r-10=180$, $5r=190$, $r = 38$. Since $\overrightarrow{BE}$ bisects $\angle ABF$, $m\angle EBF=m\angle ABE=2\times38-20=56^{\circ}$.

Answer:

1. The vertex of $\angle1$ is $A$.
2. The sides of $\angle4$ are $\overrightarrow{CA}$ and $\overrightarrow{CD}$.
3. Another name for $\angle3$ is $\angle EDA$.
4. Another name for $\angle CAD$ is $\angle 4$.
5. $m\angle FLH = 66^{\circ}$.
6. $m\angle ABE=27^{\circ}$.
7. $m\angle EBH=78^{\circ}$.
8. $m\angle EBF=16^{\circ}$.
9. $m\angle HBC=61^{\circ}$.
10. $m\angle EBF=47^{\circ}$.
11. $m\angle EBF=56^{\circ}$.