Question

in the diagram below, $overline{dg}perpoverline{df}$. use the diagram for questions 1 - 7.

1. name the vertex of $angle2$.
2. give another name for $angle3$.
3. classify $angle5$.
4. classify $angle cde$.
5. if $mangle5 = 42^{circ}$ and $mangle1 = 117^{circ}$, find $mangle cdf$.
6. if $mangle3 = 73^{circ}$, find $mangle fde$.

in the diagram below, $overline{bc}$ bisects $angle fbe$. use the diagram for questions 8 - 10.

1. if $mangle abf=(7x + 20)^{circ}$, $mangle fbc=(2x - 5)^{circ}$, and $mangle abc = 159^{circ}$, find the value of $x$.
2. if $mangle dbc=(12x - 3)^{circ}$, $mangle dbe=(5x + 12)^{circ}$, and $mangle ebc=(3x + 13)^{circ}$, find $mangle ebc$.
3. if $mangle fbc=(10x - 9)^{circ}$, $mangle cbe=(4x + 15)^{circ}$, find $mangle fbe$.

Explanation:

Step1: Recall vertex definition

The vertex of an angle is the common - point of the two rays that form the angle. For $\angle2$, the vertex is $D$.

Step2: Rename the angle

$\angle3$ can also be named $\angle GDE$ as the angle is formed by rays $\overrightarrow{DG}$ and $\overrightarrow{DE}$ with vertex $D$.

Step3: Classify $\angle5$

An acute angle is an angle whose measure is between $0^{\circ}$ and $90^{\circ}$. Since no measure is given but based on appearance and general knowledge, $\angle5$ is an acute angle.

Step4: Classify $\angle CDE$

A straight - angle has a measure of $180^{\circ}$. $\angle CDE$ is a straight - angle as the rays $\overrightarrow{CD}$ and $\overrightarrow{DE}$ form a straight line.

Step5: Find $m\angle CDF$

$\angle CDF=\angle5+\angle1$. Given $m\angle5 = 42^{\circ}$ and $m\angle1=117^{\circ}$, then $m\angle CDF=42^{\circ}+117^{\circ}=159^{\circ}$.

Step6: Find $m\angle FDE$

Since $\overrightarrow{DG}\perp\overrightarrow{DF}$, $\angle FDG = 90^{\circ}$. If $m\angle3 = 73^{\circ}$, then $m\angle FDE=m\angle3 = 73^{\circ}$ (assuming $\angle FDE$ and $\angle3$ are the same angle).

Step7: Solve for $x$ in angle - bisector problem

Since $\overrightarrow{BC}$ bisects $\angle FBE$, then $m\angle ABF + m\angle FBC=m\angle ABC$. Given $m\angle ABF=(7x + 20)^{\circ}$, $m\angle FBC=(2x - 5)^{\circ}$, and $m\angle ABC = 159^{\circ}$, we have the equation $(7x + 20)+(2x - 5)=159$. Combine like terms: $9x+15 = 159$. Subtract 15 from both sides: $9x=144$. Divide by 9: $x = 16$.

Step8: Solve for $x$ and $m\angle EBC$

Since $m\angle DBC=m\angle DBE + m\angle EBC$, we have the equation $(12x - 3)=(5x + 12)+(3x + 13)$. Combine like terms: $12x-3=8x + 25$. Subtract $8x$ from both sides: $4x-3=25$. Add 3 to both sides: $4x=28$. Divide by 4: $x = 7$. Then $m\angle EBC=(3x + 13)^{\circ}=(3\times7 + 13)^{\circ}=34^{\circ}$.

Step9: Find $m\angle FBE$

Since $\overrightarrow{BC}$ bisects $\angle FBE$, $m\angle FBE=m\angle FBC + m\angle CBE$. Given $m\angle FBC=(10x - 9)^{\circ}$ and $m\angle CBE=(4x + 15)^{\circ}$, then $m\angle FBE=(10x - 9)+(4x + 15)=14x + 6$.

Answer:

1. $D$
2. $\angle GDE$
3. Acute angle
4. Straight - angle
5. $159^{\circ}$
6. $73^{\circ}$
7. $x = 16$
8. $m\angle EBC = 34^{\circ}$
9. $m\angle FBE=14x + 6$