Question

. in this figure, ( mangle1 = 4x + 50 ) and ( mangle3 = 2x + 66 ). show all work for parts a, b, and c.

1. given: ( overrightarrow{ae}perpoverrightarrow{ab} ), ( overrightarrow{ad} ) bisects ( angle eac ), ( mangle cab = 2x - 4 ) and ( mangle cae = 3x + 14 ). show all work for parts a and b.

Explanation:

Problem 25

a. What is the value of \(x\)?

Step1: Use vertical angles property

Vertical angles are equal. \(m\angle1 = m\angle3\)
\(4x + 50=2x + 66\)

Step2: Solve for \(x\)

Subtract \(2x\) from both sides: \(4x-2x + 50=2x-2x + 66\)
\(2x+50 = 66\)
Subtract \(50\) from both sides: \(2x+50 - 50=66 - 50\)
\(2x=16\)
Divide both sides by \(2\): \(x=\frac{16}{2}=8\)

b. What is the measure of \(m\angle3\)?

Step1: Substitute \(x = 8\) into \(m\angle3\) formula

\(m\angle3=2x + 66\)
\(m\angle3=2\times8+66\)

Step2: Calculate

\(m\angle3 = 16+66=82^{\circ}\)

c. What is the measure of \(m\angle2\)?

Step1: Use linear - pair property

\(m\angle2+m\angle3 = 180^{\circ}\) (linear pair of angles)
\(m\angle2=180 - m\angle3\)

Step2: Substitute \(m\angle3 = 82^{\circ}\)

\(m\angle2=180 - 82=98^{\circ}\)

Step1: Use the fact that \(\angle EAB = 90^{\circ}\)

\(\angle EAC+\angle CAB=90^{\circ}\) (since \(\overrightarrow{AE}\perp\overrightarrow{AB}\))
\((3x + 14)+(2x-4)=90\)

Step2: Combine like terms

\(3x+2x+14 - 4=90\)
\(5x + 10=90\)

Step3: Solve for \(x\)

Subtract \(10\) from both sides: \(5x+10 - 10=90 - 10\)
\(5x=80\)
Divide both sides by \(5\): \(x=\frac{80}{5}=16\)

b. What is the measure of \(m\angle DAC\)?

Step1: Since \(AD\) bisects \(\angle EAC\)

\(m\angle DAC=\frac{1}{2}m\angle EAC\)
First, find \(m\angle EAC\): \(m\angle EAC=3x + 14\)
Substitute \(x = 16\): \(m\angle EAC=3\times16+14=48 + 14=62^{\circ}\)

Step2: Calculate \(m\angle DAC\)

\(m\angle DAC=\frac{1}{2}\times62 = 31^{\circ}\)

Answer:

a. \(x = 8\)
b. \(m\angle3=82^{\circ}\)
c. \(m\angle2=98^{\circ}\)

Problem 26

a. What is the value of \(x\)?