Question

a) write another name for ∠cbf.
b) name the sides of ∠ebd.
c) classify ∠abc.
d) give an example of an obtuse angle.
e) name two congruent angles.
f) name a perpendicular bisector.
a) name the vertex of ∠2.
b) name the sides of ∠4.
c) write another name for ∠3.
d) write another name for ∠1.
e) classify ∠ytz.
f) classify ∠ytu.
g) classify ∠xtu.
h) classify ∠wtx.
i) name two perpendicular lines.
j) name an angle bisector.
use the diagram below to answer questions 1 and 2.

1. if m∠abd = 48° and m∠dbc = 78°, find m∠abc.
2. if m∠dbc = 74° and m∠abc = 119°, find m∠abd.
3. if m∠pqr = 141°, find each measure.

(13x + 4)° (10x - 1)°
x =
m∠pqs =
m∠sqr =

1. if m∠def=(7x + 4)°, m∠deg=(5x + 1)°, and m∠gef = 23°, find each measure.

x =
m∠deg =
m∠def =

1. if m∠jkm = 43°, m∠mkl=(8x - 20)°, and m∠jkl=(10x - 11)°, find each measure.

x =
m∠mkl =

Explanation:

Step1: Recall angle - addition postulate

The measure of an angle formed by two adjacent angles is the sum of the measures of the two adjacent angles. For example, if \(\angle ABC\) is composed of \(\angle ABD\) and \(\angle DBC\), then \(m\angle ABC=m\angle ABD + m\angle DBC\).

Step2: Solve for \(m\angle ABC\) in the first - part of question 1

Given \(m\angle ABD = 48^{\circ}\) and \(m\angle DBC=78^{\circ}\), by the angle - addition postulate, \(m\angle ABC=m\angle ABD + m\angle DBC=48^{\circ}+78^{\circ}=126^{\circ}\).

Step3: Solve for \(m\angle ABD\) in the second - part of question 1

Given \(m\angle DBC = 74^{\circ}\) and \(m\angle ABC = 119^{\circ}\), using the angle - addition postulate \(m\angle ABC=m\angle ABD + m\angle DBC\), we can find \(m\angle ABD\) as \(m\angle ABD=m\angle ABC - m\angle DBC=119^{\circ}-74^{\circ}=45^{\circ}\).

Step4: Solve for \(x\) in question 3

Since \(m\angle PQR=m\angle PQS + m\angle SQR\) and \(m\angle PQR = 141^{\circ}\), \(m\angle PQS=(13x + 4)^{\circ}\), \(m\angle SQR=(10x-1)^{\circ}\), we have the equation \((13x + 4)+(10x-1)=141\).
Combining like terms: \(13x+10x+4 - 1=141\), \(23x+3 = 141\).
Subtract 3 from both sides: \(23x=141 - 3=138\).
Divide both sides by 23: \(x = 6\).

Step5: Find \(m\angle PQS\) and \(m\angle SQR\) in question 3

Substitute \(x = 6\) into the expressions for \(m\angle PQS\) and \(m\angle SQR\).
\(m\angle PQS=13x + 4=13\times6+4=82^{\circ}\).
\(m\angle SQR=10x-1=10\times6-1=59^{\circ}\).

Step6: Solve for \(x\) in question 4

Since \(m\angle DEF=m\angle DEG + m\angle GEF\), given \(m\angle DEF=(7x + 4)^{\circ}\), \(m\angle DEG=(5x + 1)^{\circ}\), and \(m\angle GEF = 23^{\circ}\), we have the equation \((7x + 4)=(5x + 1)+23\).
Expand the right - hand side: \(7x+4=5x + 1+23\), \(7x+4=5x+24\).
Subtract \(5x\) from both sides: \(7x-5x+4=5x-5x + 24\), \(2x+4=24\).
Subtract 4 from both sides: \(2x=24 - 4=20\).
Divide both sides by 2: \(x = 10\).

Step7: Find \(m\angle DEG\) and \(m\angle DEF\) in question 4

Substitute \(x = 10\) into the expressions for \(m\angle DEG\) and \(m\angle DEF\).
\(m\angle DEG=5x + 1=5\times10+1=51^{\circ}\).
\(m\angle DEF=7x + 4=7\times10+4=74^{\circ}\).

Step8: Solve for \(x\) in question 5

Since \(m\angle JKL=m\angle JKM + m\angle MKL\), given \(m\angle JKM = 43^{\circ}\), \(m\angle MKL=(8x - 20)^{\circ}\), and \(m\angle JKL=(10x-11)^{\circ}\), we have the equation \((10x-11)=43+(8x - 20)\).
Expand the right - hand side: \(10x-11=43+8x-20\), \(10x-11=8x + 23\).
Subtract \(8x\) from both sides: \(10x-8x-11=8x-8x + 23\), \(2x-11=23\).
Add 11 to both sides: \(2x=23 + 11=34\).
Divide both sides by 2: \(x = 17\).

Step9: Find \(m\angle MKL\) in question 5

Substitute \(x = 17\) into the expression for \(m\angle MKL\).
\(m\angle MKL=8x-20=8\times17-20=116^{\circ}\).

Answer:

1. \(m\angle ABC = 126^{\circ}\)
2. \(m\angle ABD = 45^{\circ}\)
3. \(x = 6\), \(m\angle PQS = 82^{\circ}\), \(m\angle SQR = 59^{\circ}\)
4. \(x = 10\), \(m\angle DEG = 51^{\circ}\), \(m\angle DEF = 74^{\circ}\)
5. \(x = 17\), \(m\angle MKL = 116^{\circ}\)