1. **Problem:** Find the missing angle or side length in a rhombus given algebraic expressions. 2. **Important properties of a rhombus:** - All sides are equal in length. - Opposite angles are equal. - Adjacent angles are supplementary (sum to 180°). - Diagonals bisect each other at right angles. 3. **Step 1: Solve for the first problem (angle at B in rhombus SBPQ with \(\angle Q = 80^\circ\))** - Since opposite angles are equal, \(\angle B = \angle Q = 80^\circ\). 4. **Step 2: Find \(m\angle T\) in rhombus UST with diagonal segments \(12x - 5\) and angle \(14x\)** - Diagonals bisect each other at right angles, so \(m\angle T = 90^\circ\). 5. **Step 3: Find \(m\angle RST\) in rhombus RSTQ with sides \(ST = 5x - 5\) and \(RQ = 5x + 10\)** - Since all sides are equal, set \(5x - 5 = 5x + 10\). - This leads to \(-5 = 10\), which is impossible, so check if sides are adjacent or opposite. - If adjacent, use supplementary angles: \(m\angle RST + m\angle STQ = 180^\circ\). - Without more info, cannot solve. 6. **Step 4: Find \(x\) in rhombus LMKJ with sides \(LM = 1 + 5x\) and \(KL = -3 + 6x\)** - Since all sides equal, set \(1 + 5x = -3 + 6x\). - Solve: \(1 + 5x = -3 + 6x\) - \(1 + 5x - 6x = -3\) - \(1 - x = -3\) - \(-x = -4\) - \(x = 4\) 7. **Step 5: Find \(x\) in rhombus KJLM with sides \(JK = 2x + 20\) and \(ML = x + 20\)** - Set equal: \(2x + 20 = x + 20\) - \(2x - x = 20 - 20\) - \(x = 0\) **Final answers:** - \(\angle B = 80^\circ\) - \(m\angle T = 90^\circ\) - \(x = 4\) in LMKJ - \(x = 0\) in KJLM Note: Problem 3 lacks sufficient info to solve.