Question

question 1-10 in the figure \\(\overrightarrow{cd}\\) and \\(\overrightarrow{cb}\\) are opposite rays, and \\(\overrightarrow{ca}\\) bisects \\(\angle bce\\). suppose \\(m\angle eca=(14x - 2)^{circ}\\) and \\(m\angle acb=(12x + 8)^{circ}). what is \\(m\angle dce?\\) \\(m\angle dce =\\)

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{CA}$ bisects $\angle BCE$, then $m\angle ECA=m\angle ACB$. So, $14x - 2=12x + 8$.

Step2: Solve for $x$

Subtract $12x$ from both sides: $14x-12x - 2=12x-12x + 8$, which gives $2x-2 = 8$. Then add 2 to both sides: $2x-2 + 2=8 + 2$, so $2x=10$. Divide both sides by 2: $x = 5$.

Step3: Find $m\angle ECB$

$m\angle ECB=m\angle ECA+m\angle ACB$. Substitute $x = 5$ into the expressions for $m\angle ECA$ and $m\angle ACB$. $m\angle ECA=14\times5-2=68^{\circ}$ and $m\angle ACB=12\times5 + 8=68^{\circ}$. So $m\angle ECB=68^{\circ}+68^{\circ}=136^{\circ}$.

Step4: Find $m\angle DCE$

Since $\overrightarrow{CD}$ and $\overrightarrow{CB}$ are opposite rays, $\angle BCD = 180^{\circ}$. Then $m\angle DCE=180^{\circ}-m\angle ECB$. Substitute $m\angle ECB = 56^{\circ}$ into the formula, $m\angle DCE=180 - 56=124^{\circ}$.

Answer:

$124$