Question

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

Explanation:

Step1: Find the measure of \(\angle AEC\)

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

$$ \angle AEC = 180^{\circ}- 72^{\circ}=108^{\circ} $$

Step2: Use the angle - bisector property

Because \(\overrightarrow{EB}\) bisects \(\angle AEC\), \(\angle AEB=\angle BEC\). Let \(\angle AEB=(7x - 2)^{\circ}\) and \(\angle BEC=(7x - 2)^{\circ}\), and \(\angle AEB+\angle BEC=\angle AEC\).

$$ (7x - 2)+(7x - 2)=108 $$

$$ 14x-4 = 108 $$

$$ 14x=108 + 4 $$

$$ 14x=112 $$

$$ x=\frac{112}{14}=8 $$

Step3: Calculate \(\angle BEC\)

Substitute \(x = 8\) into \(\angle BEC=(7x - 2)^{\circ}\)

$$ \angle BEC=(7\times8-2)^{\circ}=(56 - 2)^{\circ}=54^{\circ} $$

Answer:

\(54^{\circ}\)