Question

integrated math ii unit 1 practice test

1. using the diagram below, find all angles congruent to the given angle given l||m

linear pairs are supplementary
vertical angles are congruent
the angles of a triangle must have a sum of 180
∠4≅∠ _,∠ _,∠ _

1. given the diagram below, find y and z.

a. solve for y.
b. find m∠aec.
c. find m∠ceb.
d. solve for z.

Explanation:

Step1: Use vertical - angle property

Vertical angles are congruent. So, \(3y + 25=9y - 11\).

Step2: Solve the equation for \(y\)

Subtract \(3y\) from both sides: \(25 = 6y-11\). Then add 11 to both sides: \(36 = 6y\). Divide both sides by 6, we get \(y = 6\).

Step3: Find \(m\angle AEC\)

Substitute \(y = 6\) into the expression for \(\angle AEC\) which is \(3y + 25\). So \(m\angle AEC=3\times6 + 25=18 + 25 = 43^{\circ}\).

Step4: Find \(m\angle CEB\)

\(\angle AEC\) and \(\angle CEB\) are a linear - pair, so \(m\angle CEB=180 - m\angle AEC\). Then \(m\angle CEB = 180-43=137^{\circ}\).

Step5: Use vertical - angle property for \(z\)

Assume \((12z - 7)^{\circ}\) is vertical to either \(\angle AEC\) or \(\angle CEB\). If it is vertical to \(\angle AEC\), then \(12z-7 = 43\). Add 7 to both sides: \(12z=50\), and \(z=\frac{25}{6}\). If it is vertical to \(\angle CEB\), \(12z - 7=137\), add 7 to both sides: \(12z = 144\), then \(z = 12\). Let's assume \((12z - 7)^{\circ}\) is vertical to \(\angle AEC\) (since the problem doesn't give more context about the relationship of \(z\) - angle with others clearly). So \(z=\frac{25}{6}\).

Answer:

a. \(y = 6\)
b. \(m\angle AEC = 43^{\circ}\)
c. \(m\angle CEB = 137^{\circ}\)
d. \(z=\frac{25}{6}\)