Question

select the correct answer from each drop-down menu.
abcd is a parallelogram.
m∠x =
m∠y =
m∠z =
45
55
63
72

Explanation:

Step 1: Find \(m\angle x\)

In parallelogram \(ABCD\), opposite angles are equal. Since \( \angle C = 63^\circ\), \( \angle A=\angle C\). Thus, \(m\angle x=\angle A = 63^\circ\).

Step 2: Find \(m\angle y\)

In \(\triangle ABD\), the sum of interior angles is \(180^\circ\). We know \( \angle BAD=x = 63^\circ\) and \( \angle ABD = 72^\circ\). Using the angle - sum property of a triangle (\( \angle BAD+\angle ABD+\angle ADB = 180^\circ\)), we substitute the known values: \(63^\circ+72^\circ+m\angle y = 180^\circ\). Solving for \(m\angle y\), we get \(m\angle y=180^\circ-(63^\circ + 72^\circ)=45^\circ\).

Step 3: Find \(m\angle z\)

In parallelogram \(ABCD\), consecutive angles are supplementary. So \( \angle A+\angle D = 180^\circ\). Since \( \angle A = 63^\circ\), \( \angle D=180^\circ - 63^\circ = 117^\circ\). Also, \( \angle D=m\angle y+m\angle z\) (from the diagram, \( \angle D\) is composed of \( \angle y\) and \( \angle z\)). We know \(m\angle y = 45^\circ\), so \(m\angle z=\angle D - m\angle y=117^\circ - 45^\circ = 72^\circ\).

Answer:

\( m\angle x = 63^\circ \), \( m\angle y = 72^\circ \), \( m\angle z = 45^\circ \) (Wait, correction: Let's re - solve properly)

First, in parallelogram \(ABCD\):

Step 1: Find \(m\angle x\)

In a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Also, \(AB\parallel CD\) and \(AD\parallel BC\). Since \( \angle C = 63^\circ\), and \( \angle A=\angle C\) (opposite angles of a parallelogram), so \(m\angle x = 63^\circ\).

Step 2: Find \(m\angle y\)

In triangle \(ABD\), we know that the sum of angles in a triangle is \(180^\circ\). We know \( \angle ABD = 72^\circ\) and \( \angle BAD=x = 63^\circ\). So \(m\angle y=180-(72 + 63)=45^\circ\)? Wait, no, wait. Wait, \(AD\parallel BC\), so \( \angle ADB=\angle DBC\)? Wait, maybe I made a mistake. Let's start over.

In parallelogram \(ABCD\), \(AD\parallel BC\), so \( \angle A+\angle B = 180^\circ\), but also, \(AB\parallel CD\), so \( \angle A=\angle C = 63^\circ\) (opposite angles of parallelogram). So \(m\angle x = 63^\circ\) (since \( \angle A=x\)).

Now, in triangle \(ABD\), we have \( \angle ABD = 72^\circ\), \( \angle BAD = 63^\circ\), so \(m\angle y=180-(72 + 63)=45^\circ\)? No, wait, maybe \( \angle y\) is equal to \( \angle ABD\)? No, \(AD\parallel BC\), so \( \angle ADB=\angle DBC\), but maybe another approach. Wait, the sum of angles in a triangle is \(180^\circ\). So in triangle \(ABD\): \( \angle BAD=x\), \( \angle ABD = 72^\circ\), \( \angle ADB = y\). So \(x + 72+y=180\). We know \(x = 63\) (from parallelogram, \( \angle A=\angle C = 63\)), so \(63+72 + y=180\), \(y=180 - 63 - 72 = 45\). Then \(z = 180 - y\)? Wait, no, \( \angle y+ \angle z=180\)? No, \( \angle ADC=\angle B\), and \( \angle ADC=y + z\). Also, \( \angle B = 180-\angle A=180 - 63 = 117^\circ\), and \( \angle ABD = 72^\circ\), so \( \angle DBC=117 - 72 = 45^\circ\). Since \(AD\parallel BC\), \( \angle ADB=\angle DBC = 45^\circ\) (alternate interior angles), so \(y = 45^\circ\). Then \(z=180 - 45=135\)? No, this is getting confusing. Wait, the drop - down menu has options 45, 55, 63, 72. Let's re - examine the diagram.

Wait, the diagram shows \( \angle ABD = 72^\circ\), \( \angle C = 63^\circ\). Since \(ABCD\) is a parallelogram, \(AB\parallel CD\), so \( \angle A+\angle D = 180^\circ\), \( \angle B+\angle C = 180^\circ\). Also, \( \angle A=\angle C = 63^\circ\), so \(m\angle x = 63^\circ\).

Now, in triangle \(ABD\), angles sum to \(180^\circ\): \( \angle x=63^\circ\), \( \angle ABD = 72^\circ\), so \( \angle y=180-(63 + 72)=45^\circ\). Then \( \angle z\) and \( \angle y\) are supplementary? No, \( \angle y+ \angle z = 180\)? No, \( \angle z\) is adjacent to \( \angle y\) at point \(D\). Wait, maybe \( \angle z = 72^\circ\)? No, let's check the sum of angles in the parallelogram. The sum of angles in a parallelogram is \(360^\circ\), so \( \angle A+\angle B+\angle C+\angle D = 360\), \( \angle A=\angle C = 63\), so \( \angle B+\angle D=360 - 2\times63 = 234\), and \( \angle B=\angle D\)? No, \( \angle B+\angle D = 180\times2 - 2\times63=234\)? No, no, in a parallelogram, consecutive angles are supplementary, so \( \angle A+\angle B = 180\), \( \angle B+\angle C = 180\), etc. So \( \angle B = 180 - 63=117\), and \( \angle D = 117\). \( \angle D=y + z\), so \(y + z=117\). We found \(y = 45\), so \(z=117 - 45 = 72\)? Wait, that makes sense. So:

• \(m\angle x = 63^\circ\) (opposite angle of \( \angle C\) in parallelogram)
• \(m\angle y = 45^\circ\) (from triangle angle sum: \(63 + 72+y = 180\), \(y = 45\))
• \(m\angle z=117 - 45 = 72^\circ\) (since \( \angle D=y + z = 117\))

Wait, but the drop - down has 45, 55, 63, 72. So:

\(m\angle x = 63^\circ\) (matches the option 63)

\(m\angle y = 45^\circ\) (matches the option 45)

\(m\angle z = 72^\circ\) (matches the option 72)