Questions: If triangle ACB is congruent to triangle DCE, angle ABC=61 degrees, angle BCA=57 degrees, and angle CDE=2x degrees, x=[?]

Transcript text: If $\triangle A C B \cong \triangle D C E, \angle A B C=61^{\circ}$, $\angle B C A=57^{\circ}$, and $\angle C D E=2 x$ ${ }^{\circ}$ \[ x=[?] \]

Solution

To solve for $ x $ in the given problem, we need to use the properties of congruent triangles and the fact that the sum of the angles in a triangle is always $ 180^\circ $.

Given:

$\triangle ACB \cong \triangle DCE$
$\angle ABC = 61^\circ$
$\angle BCA = 57^\circ$
$\angle CDE = 2x^\circ$

Since $\triangle ACB \cong \triangle DCE$, corresponding angles are equal. Therefore, $\angle ACB = \angle DCE$.

First, we need to find $\angle CAB$ in $\triangle ACB$: \[ \angle CAB = 180^\circ - \angle ABC - \angle BCA \] \[ \angle CAB = 180^\circ - 61^\circ - 57^\circ \] \[ \angle CAB = 62^\circ \]

Since $\triangle ACB \cong \triangle DCE$, $\angle CAB = \angle CDE$. Therefore: \[ \angle CDE = 62^\circ \]

Given that $\angle CDE = 2x$: \[ 2x = 62^\circ \] \[ x = \frac{62^\circ}{2} \] \[ x = 31^\circ \]

Thus, the value of $ x $ is: \[ x = 31 \]

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