Question

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

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{CA}$ bisects $\angle BCE$, then $m\angle ECA=m\angle ACB$. So we set up the equation $14x - 2=12x + 8$.
$14x-12x=8 + 2$
$2x=10$
$x = 5$.

Step2: Find $m\angle ECA$

Substitute $x = 5$ into the expression for $m\angle ECA$.
$m\angle ECA=14x-2=14\times5 - 2=70 - 2=68^{\circ}$.

Step3: Find $m\angle BCE$

Since $m\angle BCE=m\angle ECA+m\angle ACB$ and $m\angle ECA=m\angle ACB = 68^{\circ}$, then $m\angle BCE=2\times68^{\circ}=136^{\circ}$.

Step4: Find $m\angle DCE$

Since $\overrightarrow{CD}$ and $\overrightarrow{CB}$ are opposite rays, $\angle BCD = 180^{\circ}$.
$m\angle DCE=180^{\circ}-m\angle BCE$.
$m\angle DCE=180 - 136=44^{\circ}$.

Answer:

$44$