Question

pass geometry a module 2 assessment
find the measures of each angle and determine whether the angles are similar.
$m\angle a =$
$m\angle b =$
$m\angle c =$
$m\angle e =$
$\triangle abc\sim\triangle dec$:

Explanation:

Step1: Identify vertical angles

$\angle ACB$ and $\angle DCE$ are vertical - angles. So, $m\angle C = 180^{\circ}-35^{\circ}-35^{\circ}=110^{\circ}$ (since the sum of angles in $\triangle DEC$ is $180^{\circ}$ and $\angle D = 35^{\circ}$ and the two non - labeled angles at $C$ are equal).

Step2: Find angles in $\triangle ABC$

In $\triangle ABC$, since $\angle ACB=\angle DCE = 110^{\circ}$, and assuming $\triangle ABC$ is isosceles (from the equal - side markings), $\angle A=\angle B$. Using the angle - sum property of a triangle ($\angle A+\angle B+\angle C = 180^{\circ}$), we have $2\angle A+110^{\circ}=180^{\circ}$. So, $\angle A=\angle B=\frac{180^{\circ}-110^{\circ}}{2}=35^{\circ}$.

Step3: Find $\angle E$ in $\triangle DEC$

In $\triangle DEC$, we know $\angle D = 35^{\circ}$ and $\angle C = 110^{\circ}$, so $\angle E=180^{\circ}-35^{\circ}-110^{\circ}=35^{\circ}$.

Step4: Determine similarity

Since $\angle A=\angle D = 35^{\circ}$, $\angle B=\angle E = 35^{\circ}$, and $\angle C=\angle C$ (vertical angles), by the AA (angle - angle) similarity criterion, $\triangle ABC\sim\triangle DEC$.

Answer:

$m\angle A = 35^{\circ}$, $m\angle B = 35^{\circ}$, $m\angle C = 110^{\circ}$, $m\angle E = 35^{\circ}$, $\triangle ABC\sim\triangle DEC$ (Yes)