1. **State the problem:** We are given trapezium PQRS with angles at S and R expressed as $(3x - 24)^\circ$ and $(102 - x)^\circ$ respectively. Angles PQR and QPS are right angles. We need to find the value of $x$. 2. **Identify key properties:** Since PQRS is a trapezium with PS parallel to QR, the consecutive interior angles between these parallel sides are supplementary (sum to $180^\circ$). 3. **Use the supplementary angle rule:** Angles at S and R are consecutive interior angles between parallel sides PS and QR, so: $$ (3x - 24) + (102 - x) = 180 $$ 4. **Simplify the equation:** $$ 3x - 24 + 102 - x = 180 $$ $$ (3x - x) + (102 - 24) = 180 $$ $$ 2x + 78 = 180 $$ 5. **Isolate $x$:** $$ 2x = 180 - 78 $$ $$ 2x = 102 $$ 6. **Divide both sides by 2:** $$ \cancel{2}x = \cancel{2}51 $$ $$ x = 51 $$ 7. **Final answer:** The value of $x$ is **51**.