find the value of x, y, and z in the rhombus below.
answer attempt 1 out of 2
x = y = z =

Question

Accepted Answer

# Explanation:
## Step1: Use the property of a rhombus (opposite angles are equal)
In a rhombus, opposite angles are equal. So, $x - 2=62$.
$$
\begin{align*}
x-2&=62\
x&=62 + 2\
x&=64
\end{align*}
$$
## Step2: Use the property that adjacent angles in a rhombus are supplementary
Adjacent angles in a rhombus are supplementary ($A + B=180^{\circ}$). Let's find $y$.
First, find the measure of the angle $- 2x-2$. Substitute $x = 64$ into $-2x - 2$: $-2\times64-2=-128 - 2=-130$.
The adjacent angle to $62^{\circ}$ is $-3y + 2$. Since $62+(-3y + 2)=180$ (adjacent angles are supplementary)
$$
\begin{align*}
62-3y+2&=180\
-3y&=180-(62 + 2)\
-3y&=116\
y&=- \frac{116}{3}\approx - 38.67
\end{align*}
$$
Wait, there is a mistake. Let's re - check.
Since $x = 64$, the angle $-2x-2=-2\times64-2=-130$. The adjacent angle to $-130^{\circ}$ is $-3y + 2$. And $-130+(-3y + 2)=180$ (adjacent angles are supplementary)
$$
\begin{align*}
-130-3y + 2&=180\
-3y&=180+130 - 2\
-3y&=308\
y&=-\frac{308}{3}\approx - 102.67
\end{align*}
$$
No, another mistake. The sum of adjacent angles in a rhombus is $180^{\circ}$.
We know that $x = 64$, so the angle $(x - 2)=62^{\circ}$, the angle $(-2x-2)=-2\times64-2=-130^{\circ}$
The angle $(-3y + 2)$ and $(-2x - 2)$ are adjacent. So $(-3y + 2)+(-2x - 2)=180$
Substitute $x = 64$
$$
\begin{align*}
-3y+2-2\times64 - 2&=180\
-3y-128&=180\
-3y&=180 + 128\
-3y&=308\
y&=-\frac{308}{3}
\end{align*}
$$
No, correct property: In a rhombus, opposite angles are equal. Let's assume the angles are $A=(x - 2)$, $B = 62^{\circ}$, $C=(-2x - 2)$, $D=(-3y + 2)$
We know $x-2 = 62$ (opposite angles), so $x = 64$
Also, $-2x-2=-3y + 2$ (opposite angles)
Substitute $x = 64$ into $-2x-2=-3y + 2$
$$
\begin{align*}
-2\times64-2&=-3y + 2\
-128-2-2&=-3y\
-132&=-3y\
y& = 44
\end{align*}
$$
## Step3: Check for $z$ (assuming no $z$ in the original problem, maybe a typo. If we assume the problem was mis - written and the angles are $x-2$, $62^{\circ}$, $-zx - 2$, $-3y + 2$
Since $x = 64$, and assume $-zx-2$ and $62^{\circ}$ are adjacent (sum to $180^{\circ}$)
$-zx-2+62=180$
Substitute $x = 64$
$-64z+60 = 180$
$-64z=120$
$z=-\frac{120}{64}=-\frac{15}{8}=-1.875$
If assume $-zx - 2$ and $-3y + 2$ are opposite (equal)
$-zx-2=-3y + 2$
Substitute $x = 64$ and $y = 44$
$-64z-2=-3\times44+2$
$-64z-2=-132 + 2$
$-64z-2=-130$
$-64z=-128$
$z = 2$

# Answer:
$x = 64$, $y = 44$, $z = 2$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the property of a rhombus (opposite angles are equal)

Step2: Use the property that adjacent angles in a rhombus are supplementary

Step3: Check for \(z\) (assuming no \(z\) in the original problem, maybe a typo. If we assume the problem was mis - written and the angles are \(x-2\), \(62^{\circ}\), \(-zx - 2\), \(-3y + 2\)

Answer: