1. **Problem statement:** We are given a trapezium with one side length 14.9 cm, bottom left angle 75°, bottom right angle 50°, and we need to find the length $r$ of the slant right side. 2. **Understanding the trapezium:** The trapezium has two parallel sides (top and bottom) and two non-parallel sides (left and right). We know the left side length and the two bottom angles. 3. **Approach:** We can use the Law of Sines or trigonometric relations in the triangles formed by dropping perpendiculars or by considering the trapezium as two triangles. 4. **Step 1: Calculate the top angle at the left side:** Since the trapezium's adjacent angles between the parallel sides sum to 180°, the top left angle is $180^\circ - 75^\circ = 105^\circ$. 5. **Step 2: Calculate the top angle at the right side:** Similarly, the top right angle is $180^\circ - 50^\circ = 130^\circ$. 6. **Step 3: Use the Law of Sines in the triangle formed by the left side and the top and bottom angles:** Let the trapezium be $ABCD$ with $AB$ top, $BC$ right side (length $r$), $CD$ bottom, and $DA$ left side (14.9 cm). In triangle $ABD$, angles at $A$ and $D$ are 105° and 75°, side $DA=14.9$ cm. Using Law of Sines: $$\frac{AB}{\sin 75^\circ} = \frac{14.9}{\sin 105^\circ}$$ Calculate $AB$: $$AB = \frac{14.9 \times \sin 75^\circ}{\sin 105^\circ}$$ 7. **Step 4: Calculate $AB$ numerically:** $\sin 75^\circ \approx 0.9659$, $\sin 105^\circ \approx 0.9659$ So, $$AB = \frac{14.9 \times 0.9659}{0.9659} = 14.9 \text{ cm}$$ 8. **Step 5: Use Law of Sines in triangle $BCD$ to find $r$:** Angles at $C$ and $D$ are 130° and 50°, side $CD$ is unknown but $AB$ and $CD$ are parallel. Using Law of Sines: $$\frac{r}{\sin 50^\circ} = \frac{CD}{\sin 130^\circ}$$ But $CD$ can be found from the trapezium properties or by using the fact that $AB$ and $CD$ are parallel and the trapezium's height can be found. 9. **Step 6: Calculate height $h$ of trapezium using left side and angle 75°:** $$h = 14.9 \times \sin 75^\circ = 14.9 \times 0.9659 = 14.39 \text{ cm}$$ 10. **Step 7: Calculate bottom base $CD$ using height and angle 50°:** $$CD = h / \sin 50^\circ = 14.39 / 0.7660 = 18.79 \text{ cm}$$ 11. **Step 8: Use Law of Cosines in triangle $BCD$ to find $r$:** We know $CD = 18.79$ cm, angle at $C = 130^\circ$, and height $h = 14.39$ cm. Using Law of Cosines: $$r^2 = h^2 + (CD)^2 - 2 \times h \times CD \times \cos 130^\circ$$ Calculate: $$r^2 = 14.39^2 + 18.79^2 - 2 \times 14.39 \times 18.79 \times \cos 130^\circ$$ 12. **Step 9: Calculate $r$ numerically:** $\cos 130^\circ = -0.6428$ $$r^2 = 207.07 + 353.17 + 2 \times 14.39 \times 18.79 \times 0.6428$$ Calculate the last term: $$2 \times 14.39 \times 18.79 \times 0.6428 = 347.5$$ So, $$r^2 = 207.07 + 353.17 + 347.5 = 907.74$$ $$r = \sqrt{907.74} = 30.13 \text{ cm}$$ **Final answer:** $$r = 30.13 \text{ cm}$$