Question

find the value of ( x ), ( y ), and ( z ) in the rhombus below.

answer attempt 1 out of 2
( x = square ) ( y = square ) ( z = square )
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Explanation:

Step1: Recall properties of a rhombus

In a rhombus, opposite angles are equal, and adjacent angles are supplementary (sum to \(180^\circ\)).

Step2: Solve for \(x\)

The angle \((-2x - 2)^\circ\) and \(62^\circ\) are adjacent angles? Wait, no, in a rhombus, opposite angles are equal. Wait, maybe I made a mistake. Wait, in a rhombus, opposite angles are equal, and adjacent angles are supplementary. Let's check the diagram. Let's assume that the angle \((-2x - 2)^\circ\) is opposite to... Wait, no, maybe the angle \((-2x - 2)^\circ\) and \(62^\circ\) are adjacent? Wait, no, let's think again. Wait, in a rhombus, consecutive angles are supplementary. So if one angle is \(62^\circ\), then the consecutive angle is \(180 - 62 = 118^\circ\). So \(-2x - 2 = 118\)? Wait, no, wait: \(-2x - 2 = 180 - 62\)? Wait, \(180 - 62 = 118\). So \(-2x - 2 = 118\). Let's solve for \(x\):

\(-2x - 2 = 118\)

Add 2 to both sides: \(-2x = 120\)

Divide by -2: \(x = -60\)? Wait, that can't be right. Wait, maybe the angle \((-2x - 2)^\circ\) is equal to \(62^\circ\)? Wait, no, that would give a negative \(x\). Wait, maybe I mixed up. Wait, let's check the other angle: \((-3y + 2)^\circ\). Wait, maybe \((-3y + 2)^\circ\) is equal to \(62^\circ\)? Wait, no, let's start over.

In a rhombus, opposite angles are equal. So the angle labeled \(62^\circ\) and the angle \((-3y + 2)^\circ\) are opposite? Wait, no, the diagram shows the rhombus with angles: top left: \((-2x - 2)^\circ\), top right: \(62^\circ\), bottom left: \((-3y + 2)^\circ\), bottom right: \((z - 2)^\circ\). So in a rhombus, opposite angles are equal. So top left and bottom right are opposite, top right and bottom left are opposite. So \(62^\circ = (-3y + 2)^\circ\), and \((-2x - 2)^\circ = (z - 2)^\circ\). Also, adjacent angles are supplementary. So top left + top right = \(180^\circ\), so \((-2x - 2) + 62 = 180\)? Wait, no, that would be if they are adjacent. Wait, let's confirm: in a rhombus, consecutive angles are supplementary. So if top right is \(62^\circ\), then top left (consecutive) is \(180 - 62 = 118^\circ\). So \(-2x - 2 = 118\). Let's solve:

\(-2x - 2 = 118\)

Add 2 to both sides: \(-2x = 120\)

Divide by -2: \(x = -60\). Wait, that's negative, but maybe it's correct? Wait, no, maybe I made a mistake. Wait, maybe the angle \((-2x - 2)^\circ\) is equal to \(62^\circ\)? Let's try that:

\(-2x - 2 = 62\)

Add 2: \(-2x = 64\)

Divide by -2: \(x = -32\). No, that's also negative. Wait, maybe the angle \((-3y + 2)^\circ\) is equal to \(62^\circ\). Let's solve for \(y\):

\(-3y + 2 = 62\)

Subtract 2: \(-3y = 60\)

Divide by -3: \(y = -20\). No, that's not right. Wait, maybe I got the supplementary angles wrong. Wait, in a rhombus, opposite angles are equal, and adjacent angles are supplementary. So if one angle is \(62^\circ\), the angle opposite to it is also \(62^\circ\), and the angles adjacent to it are \(180 - 62 = 118^\circ\). So let's look at the angles:

Top right: \(62^\circ\)

Bottom left: \((-3y + 2)^\circ\) – if these are opposite, then \(-3y + 2 = 62\) → \(-3y = 60\) → \(y = -20\). No, that's negative. Wait, maybe the angle \((-3y + 2)^\circ\) is adjacent to \(62^\circ\), so \((-3y + 2) + 62 = 180\) → \(-3y + 64 = 180\) → \(-3y = 116\) → \(y = -116/3\). No, that's not right. Wait, maybe the angle \((-2x - 2)^\circ\) is adjacent to \(62^\circ\), so \((-2x - 2) + 62 = 180\) → \(-2x + 60 = 180\) → \(-2x = 120\) → \(x = -60\). Then the angle \((-2x - 2) = -2*(-60) - 2 = 120 - 2 = 118^\circ\), which is supplementary to \(62^\circ\) (118 + 62 = 180), that works. Then the angle \((z - 2)^\circ\) is opposite to \((-2x - 2)^\circ\), so \(z - 2 = 118\) → \(z = 120\). Then the angle \((-3y + 2)^\circ\) is opposite to \(62^\circ\), so \(-3y + 2 = 62\) → \(-3y = 60\) → \(y = -20\). Wait, but \(y\) is negative? That seems odd, but mathematically it works. Wait, maybe the diagram is labeled differently. Let's check again.

Wait, maybe the angle \((-2x - 2)^\circ\) is equal to \(62^\circ\)? Let's try:

\(-2x - 2 = 62\)

\(-2x = 64\)

\(x = -32\)

Then the adjacent angle would be \(180 - 62 = 118\), so \(z - 2 = 118\) → \(z = 120\), and \(-3y + 2 = 118\) → \(-3y = 116\) → \(y = -116/3\). No, that's worse.

Wait, maybe the problem is that the angle \((-2x - 2)^\circ\) is supplementary to \(62^\circ\), so:

\(-2x - 2 + 62 = 180\)

\(-2x + 60 = 180\)

\(-2x = 120\)

\(x = -60\)

Then the angle \((-2x - 2) = -2*(-60) - 2 = 118^\circ\), which is correct (118 + 62 = 180). Then the angle \((z - 2)^\circ\) is equal to \((-2x - 2)^\circ\) (opposite angles), so:

\(z - 2 = 118\)

\(z = 120\)

Then the angle \((-3y + 2)^\circ\) is equal to \(62^\circ\) (opposite angles), so:

\(-3y + 2 = 62\)

\(-3y = 60\)

\(y = -20\)

So even though \(x\) and \(y\) are negative, mathematically it works because the angle measures are positive. Let's check:

For \(x = -60\): \(-2x - 2 = -2*(-60) - 2 = 120 - 2 = 118^\circ\), which is positive.

For \(y = -20\): \(-3y + 2 = -3*(-20) + 2 = 60 + 2 = 62^\circ\), which matches.

For \(z = 120\): \(z - 2 = 118^\circ\), which matches the opposite angle.

So that works.

Answer:

\(x = -60\), \(y = -20\), \(z = 120\)