Question

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

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 $14x - 2=12x + 8$.

Step2: Solve the equation for $x$

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

Step3: Find the measure of $\angle ECA$

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

Answer:

$68$