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 \(AE = EB\) (marked as equal) and \(AE\parallel BC\) (implied by the parallel - line markings), triangle \(AEC\) is isosceles (\(AC = EC\), because \(AE\parallel BC\) and \(AE = EB\) gives some congruent - angle relationships). Also, \(\angle A=\angle AEC = 62^{\circ}\) (alternate interior angles for \(AE\parallel BC\) are not relevant here, but for the line - angle relationship, \(\angle AEC+(11x - 2)^{\circ}=180^{\circ}\), and \(\angle EBC=(6x + 13)^{\circ}\). Since \(AE = EB\), \(\angle ACE=\angle ECB\) (by the property of the line \(CE\) in the isosceles - like structure formed by \(AC = EC\)).
First, use the angle - sum property of a triangle. In triangle \(ABC\), we know that \(\angle A = 62^{\circ}\). Also, since \(AE = EB\) and \(AE\parallel BC\) (by the mid - point theorem and parallel - line angle relationships), we can find \(x\) from the equation \(11x-2=180 - 62\) (not the right approach). Wait, correct approach:
Since \(AE\parallel BC\), \(\angle A=\angle ECB\) (alternate interior angles) is wrong. Wait, correct:
Since \(AC = EC\) (because \(AE = EB\) and \(AE\parallel BC\) gives some congruent - triangle or isosceles - triangle properties). Let's use the angle - sum in the large - triangle - related concept.
We know that \(\angle A = 62^{\circ}\). Also, since \(AE = EB\) and \(AE\parallel BC\) (by the mid - segment theorem and angle relationships), we can use the fact that \(\angle AEC+(11x - 2)^{\circ}=180^{\circ}\) (linear pair). But another way:
Since \(AE = EB\) and \(AE\parallel BC\) (by the mid - point theorem and parallel - line properties), we can use the angle - sum in triangle \(ABC\).
We know that \(\angle A = 62^{\circ}\). Also, since \(AE = EB\) and \(AE\parallel BC\) (by the mid - segment theorem and parallel - line properties), we can use the fact that \(\angle ACE=\angle ECB\).
First, find \(x\):
Since \(AE = EB\), \(\angle ACE=\angle ECB\) (by the property of the line \(CE\) in the isosceles - like structure formed by \(AC = EC\)).
We know that \(\angle A = 62^{\circ}\). Also, using the angle - sum property of a triangle \(180^{\circ}=\angle A+\angle ABC+\angle ACB\).
Since \(AE = EB\) and \(AE\parallel BC\) (by the mid - point theorem and parallel - line properties), \(\angle AEC=(11x - 2)^{\circ}\) and \(\angle EBC=(6x + 13)^{\circ}\). Also, \(\angle AEC+\angle EBC = 180^{\circ}-\angle A\) (by the exterior - angle property or parallel - line - related angle sum).
Wait, correct:
Since \(AE = EB\) and \(AE\parallel BC\) (by the mid - point theorem and parallel - line properties), we use the fact that \(\angle AEC=(11x - 2)^{\circ}\) and \(\angle EBC=(6x + 13)^{\circ}\). Also, \(\angle AEC=\angle EBC+\angle ECB\) (exterior - angle property of triangle \(EBC\)). But another approach:
Since \(AE = EB\) (given), \(\angle ACE=\angle ECB\) (by the property of the line \(CE\) in the isosceles - like structure formed by \(AC = EC\)).
We know that \(\angle A = 62^{\circ}\). Also, using the angle - sum property of a triangle \(180^{\circ}=\angle A+\angle ABC+\angle ACB\).
Since \(AE = EB\) and \(AE\parallel BC\) (by the mid - point theorem and parallel - line properties), we can find \(x\) from \(11x-2=6x + 13+62\) (using the exterior - angle property of triangle \(ABC\) where \(\angle AEC\) (with measure \((11x - 2)^{\circ}\)) is an exterior angle for triangle \(EBC\) and \(\angle A\) is related).
\[ \begin{align*} 11x-2&=6x + 13+62\\ 11x-6x&=13 + 62+2\\ 5x&=77\\ x& = 11 \end{align*} \]

Step2: Calculate \(\angle ACB\)

Substitute \(x = 11\) into the expression for \(\angle ACB\).
We know that \(\angle ACB=180-(62+(6x + 13))\) (using angle - sum of triangle \(ABC\)).
First, \(\angle ABC=(6x + 13)^{\circ}\), substituting \(x = 11\), \(\angle ABC=(6\times11 + 13)^{\circ}=(66 + 13)^{\circ}=79^{\circ}\)
Then \(\angle ACB=180-(62 + 79)=39^{\circ}\)

Answer:

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