Question

10 fill in the blank 1 point find ( m angle a c b ). navigate to question at position 10 ( x = ) ( m angle a c b = )

Explanation:

Step1: Use the property of isosceles triangle and parallel lines

Since \(AD = DC\) and \(DE\parallel AB\), \(\angle A=\angle ADE = 62^{\circ}\), and \(\angle ADE=\angle DEC = 62^{\circ}\) (alternate - interior angles). Also, \(\angle DEC+\angle DEB = 180^{\circ}\) (linear - pair), so \(\angle DEB=180 - 62=118^{\circ}\).
Since \(DE\parallel AB\), \(\angle DEB+\angle B = 180^{\circ}\) (same - side interior angles). And \(\angle AED=(11x - 2)^{\circ}\), \(\angle B=(6x + 13)^{\circ}\). Also, because \(DE\parallel AB\), \(\angle AED+\angle B = 180^{\circ}\) (same - side interior angles).
So, \((11x - 2)+(6x + 13)=180\).

Step2: Solve the equation for \(x\)

Combine like terms: \(11x+6x-2 + 13=180\), \(17x+11 = 180\).
Subtract \(11\) from both sides: \(17x=180 - 11=169\).
\(x = 11\).

Step3: Use the property of mid - segment and triangle angle - sum

Since \(DE\) is the mid - segment of \(\triangle ABC\) (\(AD = DC\) and \(DE\parallel AB\)), \(CE=EB\).
In \(\triangle CEB\), \(\angle B=(6x + 13)^{\circ}\), substituting \(x = 11\), \(\angle B=(6\times11 + 13)=66 + 13=79^{\circ}\).
In \(\triangle ABC\), \(\angle A = 62^{\circ}\), \(\angle B = 79^{\circ}\).
By the triangle angle - sum theorem (\(\angle A+\angle B+\angle ACB=180^{\circ}\)), \(\angle ACB=180-(62 + 79)=39^{\circ}\).

Answer:

\(m\angle ACB = 39^{\circ}\)