Question

if $delta acbcongdelta dce$, $angle abc = 61^{circ}$, $angle bca = 57^{circ}$, and $angle cde = 2x$. $x=?$

Explanation:

Step1: Find $\angle BAC$

In $\triangle ACB$, use angle - sum property of a triangle. The sum of interior angles of a triangle is $180^{\circ}$. So, $\angle BAC=180^{\circ}-\angle ABC - \angle BCA$. Substitute $\angle ABC = 61^{\circ}$ and $\angle BCA = 57^{\circ}$, then $\angle BAC=180^{\circ}-61^{\circ}-57^{\circ}=62^{\circ}$.

Step2: Use congruent - triangle property

Since $\triangle ACB\cong\triangle DCE$, corresponding angles are equal. $\angle CDE=\angle BAC$. Given $\angle CDE = 2x$ and $\angle BAC = 62^{\circ}$, we set up the equation $2x=62^{\circ}$.

Step3: Solve for $x$

Divide both sides of the equation $2x = 62^{\circ}$ by 2. We get $x=\frac{62^{\circ}}{2}=31^{\circ}$.

Answer:

$31$