In a rhombus, opposite angles are equal, and the sum of all angles in any quadrilateral is \(360^\circ\).
Given the angles:
- \(88^\circ\)
- \((2x + 10)^\circ\)
- \((-3y - 2)^\circ\)
- \((-6z + 2)^\circ\)
Step 1: Set Up Equation for Angles
Since opposite angles are equal, we know:
- \((2x + 10)^\circ = 88^\circ\)
- \((-3y - 2)^\circ = 88^\circ\)
Step 2: Solve for \(x\)
From the first equation: \[ 2x + 10 = 88 \] Subtract 10 from both sides: \[ 2x = 78 \] Now divide by 2: \[ x = 39 \]
Step 3: Solve for \(y\)
From the second equation: \[ -3y - 2 = 88 \] Add 2 to both sides: \[ -3y = 90 \] Divide by -3: \[ y = -30 \]
Step 4: Find the Fourth Angle
Now, let's find the fourth angle using the sum of the angles in a quadrilateral: \[ 88 + (2x + 10) + (-3y - 2) + (-6z + 2) = 360 \] Substituting the solved values:
- \(x = 39\) so \(2x + 10 = 2(39) + 10 = 88 + 10 = 98\)
- \(y = -30\) so \(-3y - 2 = -3(-30) - 2 = 90 - 2 = 88\)
Now, plug these in: \[ 88 + 98 + 88 + (-6z + 2) = 360 \] Combine the known angles: \[ 274 + (-6z + 2) = 360 \] Now simplify: \[ -6z + 2 = 360 - 274 \] \[ -6z + 2 = 86 \] Subtract 2 from both sides: \[ -6z = 84 \] Divide by -6: \[ z = -14 \]
Final Answers
Thus, the values are: \[ x = 39, \quad y = -30, \quad z = -14 \]