Questions: If triangle ABC is congruent to triangle DEC, angle ABC = 2x, angle BCA = 48°, and angle CDE = 74°

Transcript text: If $\triangle A B C \cong \triangle D E C, \angle A B C=2 x$, $\angle B C A=48^{\circ}$, and $\angle C D E=74^{\circ}$

Solution

Solution Steps

Step 1: Find the measure of $\angle ACB$.

Since the sum of the angles in $\triangle ABC$ is $180^\circ$, we have $\angle CAB + \angle ABC + \angle BCA = 180^\circ$. Substituting the given values, we get $\angle CAB + 2x + 48 = 180$, so $\angle CAB = 132 - 2x$.

Step 2: Use the congruent triangles.

Since $\triangle ABC \cong \triangle DEC$, we have corresponding angles congruent. Therefore, $\angle ABC \cong \angle DEC$, so $2x = \angle DEC$. Also, $\angle BCA \cong \angle ECD$, so $48^\circ = \angle ECD$. And, $\angle CAB \cong \angle CDE$, so $132 - 2x = 74$.

Step 3: Solve for x.

$132 - 2x = 74$ $2x = 132 - 74$ $2x = 58$ $x = 29$