Question

1. use the figure to identify the angles or parts of angles that satisfy each given condition.

a) name ∠1 in two different ways.
b) name the sides of ∠2.
c) name ∠cbg in two different ways.
d) name a point in the interior of ∠fde.
e) name all of the points in the exterior of ∠fde.
f) name two opposite rays.
g) classify ∠edh.
h) classify ∠cdh.
i) classify ∠abc.

1. in the figure, (overrightarrow{ba}) and (overrightarrow{bc}) are opposite rays and (overrightarrow{bd}) bisects ∠abe. if (mangle abd=(4x + 14)^{circ}) and (mangle dbe=(8x - 32)^{circ}), find (mangle dbe).
2. in the figure, (overrightarrow{kj}) and (overrightarrow{km}) are opposite rays and (overrightarrow{kn}) bisects ∠jkl. if (mangle jkn=(8x - 13)^{circ}) and (mangle jkl=(12x + 22)^{circ}), find (mangle jkn).

Explanation:

Step1: Recall angle - naming and angle - bisector properties

An angle can be named in multiple ways using the vertex and points on its rays. An angle - bisector divides an angle into two equal parts.

Step2: Solve problem 2

Since $\overrightarrow{BD}$ bisects $\angle ABE$, then $m\angle ABD=m\angle DBE$.
Set up the equation: $4x + 14=8x-32$.
Subtract $4x$ from both sides: $14 = 4x-32$.
Add 32 to both sides: $4x=14 + 32=46$.
Divide both sides by 4: $x=\frac{46}{4}=11.5$.
Substitute $x = 11.5$ into the expression for $m\angle DBE$: $m\angle DBE=8x-32=8\times11.5-32=92 - 32=60^{\circ}$.

Step3: Solve problem 3

Since $\overrightarrow{KN}$ bisects $\angle JKL$, then $m\angle JKL = 2m\angle JKN$.
Set up the equation: $12x + 22=2(8x-13)$.
Expand the right - hand side: $12x + 22=16x-26$.
Subtract $12x$ from both sides: $22=4x-26$.
Add 26 to both sides: $4x=22 + 26=48$.
Divide both sides by 4: $x = 12$.
Substitute $x = 12$ into the expression for $m\angle JKN$: $m\angle JKN=8x-13=8\times12-13=96-13=83^{\circ}$.

Answer:

1. $m\angle DBE = 60^{\circ}$
2. $m\angle JKN=83^{\circ}$