Question

1. find ( mangle acb ).

Explanation:

Step1: Use the mid - segment theorem

Since \(DE\) is the mid - segment of \(\triangle ACB\) (\(AD = DC\) and \(AE=EB\)), then \(DE\parallel CB\). So, \(\angle AED=\angle B=(6x + 13)^{\circ}\) (corresponding angles). Also, by the property of the exterior angle of a triangle, \(\angle AED\) is an exterior angle of \(\triangle ADE\). So, \(\angle AED=\angle A+\angle ADE\). But since \(DE\parallel CB\), we can also use the fact that \(\angle AED=(11x - 2)^{\circ}\) (given). And because \(DE\parallel CB\), \(\angle AED=\angle B\) (corresponding angles) and from the exterior - angle property of \(\triangle ADE\) (where \(\angle A = 62^{\circ}\)), we have \(11x-2=6x + 13\).

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Step2: Find \(\angle B\)

Substitute \(x = 3\) into \(\angle B=(6x + 13)^{\circ}\). Then \(\angle B=(6\times3+13)^{\circ}=(18 + 13)^{\circ}=31^{\circ}\).

Step3: Use the angle - sum property of a triangle

In \(\triangle ACB\), we know that \(\angle A+\angle B+\angle ACB=180^{\circ}\). Given \(\angle A = 62^{\circ}\) and \(\angle B = 31^{\circ}\).

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Answer:

\(m\angle ACB = 87^{\circ}\)