1. **Problem statement:** We are given two triangles with angles and sides, and we need to solve for $x$ in the second triangle. 2. **Identify known angles and sides:** - In the first triangle, angles are $59^\circ$, $79^\circ$, and $42^\circ$ (since the sum of angles in a triangle is $180^\circ$). - Side opposite $59^\circ$ is 16. - Side $ML$ is 11. - In the second triangle, angles are $79^\circ$, $42^\circ$, and $59^\circ$. - Side adjacent to $42^\circ$ is 46. - Side adjacent to $79^\circ$ is $x$. 3. **Use the Law of Sines:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a,b,c$ are sides opposite angles $A,B,C$ respectively. 4. **Apply Law of Sines to second triangle:** We know: - Side opposite $42^\circ$ is 46. - Side opposite $79^\circ$ is $x$. So: $$\frac{x}{\sin 79^\circ} = \frac{46}{\sin 42^\circ}$$ 5. **Solve for $x$:** $$x = \frac{46 \times \sin 79^\circ}{\sin 42^\circ}$$ 6. **Calculate values:** - $\sin 79^\circ \approx 0.9816$ - $\sin 42^\circ \approx 0.6691$ 7. **Substitute and compute:** $$x = \frac{46 \times 0.9816}{0.6691}$$ 8. **Simplify:** $$x = \frac{\cancel{46} \times 0.9816}{0.6691} = 46 \times 1.466 = 67.4$$ 9. **Final answer:** $$x \approx 67.4$$ Thus, the value of $x$ rounded to the nearest tenth is 67.4.