Question

in the figure, $overrightarrow{ed}$ and $overrightarrow{ea}$ are opposite rays, and $overrightarrow{eb}$ bisects $angle aec$. if $mangle ced = 72^{circ}$ and $mangle aeb=(7x - 2)^{circ}$, then what is $mangle bec$? $mangle bec=square^{circ}$

Explanation:

Step1: Use linear - pair property

Since $\overrightarrow{ED}$ and $\overrightarrow{EA}$ are opposite rays, $\angle AEC+\angle CED = 180^{\circ}$. Given $\angle CED=72^{\circ}$, then $\angle AEC=180^{\circ}-\angle CED = 180 - 72=108^{\circ}$.

Step2: Apply angle - bisector property

Since $\overrightarrow{EB}$ bisects $\angle AEC$, $\angle AEB=\angle BEC$. And we know that $\angle AEB=(7x - 2)^{\circ}$ and $\angle AEB=\frac{1}{2}\angle AEC$. So $\angle AEB=\frac{1}{2}\times108^{\circ}=54^{\circ}$.

Step3: Solve for $x$ (not required for final answer but for understanding)

Set $7x - 2=54$. Then $7x=54 + 2=56$, and $x = 8$.

Answer:

$54$