Question

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x
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m∠ced = 3x + 2
x=
m∠1 =

Explanation:

Step1: Identify angle relationship

$\angle CED$ and $\angle AEB$ are vertical - angles. Since $\angle AEB=x$ and $\angle CED = 3x + 2$, and vertical angles are equal, we have the equation $x=3x + 2$.

Step2: Solve the equation for x

Subtract $3x$ from both sides: $x-3x=3x + 2-3x$. This gives $-2x=2$. Then divide both sides by $-2$: $x=\frac{2}{-2}=-1$. But angles cannot have a negative measure in this context. There is a mistake above. We should note that $\angle CED$ and $\angle AEB$ are not vertical - angles. Since $\angle CED$ and $\angle AEB$ are supplementary (because $\angle CED+\angle AEB = 180^{\circ}$ as $\angle CED$ and $\angle AEB$ form a linear - pair), we have the equation $x+(3x + 2)=180$.

Step3: Simplify the new equation

Combine like terms: $x+3x+2 = 180$, which simplifies to $4x+2 = 180$.

Step4: Solve for x

Subtract 2 from both sides: $4x+2 - 2=180 - 2$, so $4x=178$. Then divide both sides by 4: $x=\frac{178}{4}=44.5$.

Step5: Find $m\angle CED$

Substitute $x = 44.5$ into the expression for $\angle CED$. $m\angle CED=3x + 2=3\times44.5+2=133.5 + 2=135.5^{\circ}$.

Answer:

$x = 44.5$
$m\angle CED=135.5^{\circ}$ (assuming $\angle1$ is $\angle CED$)