1. **Problem Statement:** Solve for $x$ in the conjoined triangles $QRU$ and $TSU$ where both have an angle of $54^\circ$ and share side $TU$. 2. **Given:** - Triangle $QRU$ with $\angle Q = 54^\circ$ and side $RU = 3$. - Triangle $TSU$ with $\angle T = 54^\circ$, sides $SU = 6$ and $TS = 7$. - Segment $QS = x$. 3. **Approach:** Since both triangles share side $TU$ and have an angle of $54^\circ$, we can use the Law of Sines to find $TU$ in each triangle and set them equal. 4. **Law of Sines formula:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 5. **In triangle $QRU$:** - Let $\angle R = \theta$ and $\angle U = 180^\circ - 54^\circ - \theta = 126^\circ - \theta$. - Using Law of Sines: $$\frac{RU}{\sin Q} = \frac{TU}{\sin R}$$ $$\Rightarrow TU = \frac{RU \cdot \sin R}{\sin Q} = \frac{3 \sin R}{\sin 54^\circ}$$ 6. **In triangle $TSU$:** - Let $\angle S = \phi$ and $\angle U = 180^\circ - 54^\circ - \phi = 126^\circ - \phi$. - Using Law of Sines: $$\frac{TS}{\sin U} = \frac{TU}{\sin T}$$ $$\Rightarrow TU = \frac{TS \cdot \sin T}{\sin U} = \frac{7 \sin 54^\circ}{\sin U}$$ 7. **Since $TU$ is common:** $$\frac{3 \sin R}{\sin 54^\circ} = \frac{7 \sin 54^\circ}{\sin U}$$ 8. **We need to find $x = QS$ which is the sum of $QR + RS$ or directly relate it via triangle properties.** 9. **Alternatively, use Law of Cosines or other given data to find $x$. However, with given data and angles equal, $x$ corresponds to $QS = SU = 6$ (since $SU$ is given as 6 and $QS$ is opposite the same angle).** 10. **Therefore, the value of $x$ is:** $$x = 6$$ **Final answer:** $$\boxed{6}$$