Question

quadrilateral qrst is a rhombus and $m\angle rqs = 3y + 32^\circ$. what is the value of y?
$y = \square ^\circ$

Explanation:

Step1: Use rhombus side property

In rhombus $QRST$, $QS$ is a side, so $QS=QT$, making $\triangle QST$ isosceles. Thus, $\angle QST = \angle QTS = 62^\circ$.

Step2: Calculate $\angle TQS$

Sum of angles in a triangle is $180^\circ$.
$\angle TQS = 180^\circ - 62^\circ - 62^\circ = 56^\circ$

Step3: Use rhombus diagonal property

Diagonals of a rhombus bisect angles. $\angle RQT = \angle TQS = 56^\circ$, and diagonal $QS$ bisects $\angle RQT$, so $\angle RQS = \frac{1}{2}\angle RQT$.
$\angle RQS = \frac{1}{2} \times 56^\circ = 28^\circ$

Step4: Solve for y

Set $3y + 32^\circ = 28^\circ$
$3y = 28^\circ - 32^\circ$
$3y = -4^\circ$
$y = -\frac{4}{3}^\circ$

Answer:

$y = -\frac{4}{3}$