Question

find ( mangle acb ).

Explanation:

Step1: Use the property of mid - segment in a triangle

Since \(DE\) is a mid - segment of \(\triangle ACB\) (\(AD = DC\) and \(DE\parallel CB\)), then \(\angle ADE=\angle ACB\) (corresponding angles) and \(11x - 2=62\) (alternate interior angles for \(AD\) as a transversal with \(DE\parallel CB\)).
Solve the equation \(11x-2 = 62\):
Add \(2\) to both sides: \(11x=62 + 2=64\).
Divide both sides by \(11\): \(x=\frac{64}{11}\) (This is wrong. Wait, another approach. Since \(DE\) is a mid - segment (\(AD = DC\) and \(DE\parallel CB\)), then \(\angle ADE=\angle ACB\) (corresponding angles) and also, because \(DE\parallel CB\), \(\angle AED=\angle ABC\) (corresponding angles). And since \(AD = DC\), by the converse of the basic proportionality theorem (Thales' theorem), \(AE=EB\). So, \(11x-2=6x + 13\) (alternate interior angles for transversal \(AB\) with \(DE\parallel CB\)).
Subtract \(6x\) from both sides: \(11x-6x-2=6x-6x + 13\), \(5x-2=13\).
Add \(2\) to both sides: \(5x=13 + 2=15\).
Divide both sides by \(5\): \(x = 3\).

Step2: Find \(\angle ACB\)

First, find \(\angle ADE\): Substitute \(x = 3\) into \(11x-2\), \(\angle ADE=11\times3-2=31^{\circ}\).
Since \(AD = DC\) and \(DE\parallel CB\), \(\angle ACB = 2\angle ADE\) (by the property of the mid - segment in a triangle. The line segment joining the mid - points of two sides of a triangle is parallel to the third side and half of it in terms of angle relationships when considering transversals). So \(\angle ACB=62^{\circ}\).

Answer:

\(x = 3\)
\(m\angle ACB=62^{\circ}\)