1. **State the problem:** We have two right triangles, ABC and ACD, with given angles and some side lengths. We need to find the missing sides $k$, $m$, $x$, and $y$. 2. **Triangle ABC:** Given angles $\angle B = 45^\circ$, $\angle A = 60^\circ$, and $\angle C = 90^\circ$, and side $AC = 6$. 3. **Use the fact that in a right triangle, the sides relate to angles by sine and cosine:** - $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ 4. **Find side $k = AB$ (hypotenuse of triangle ABC):** Since $AC$ is opposite $\angle B = 45^\circ$, $$\sin(45^\circ) = \frac{AC}{AB} = \frac{6}{k}$$ So, $$k = \frac{6}{\sin(45^\circ)}$$ Using $\sin(45^\circ) = \frac{\sqrt{2}}{2}$, $$k = \frac{6}{\frac{\sqrt{2}}{2}} = 6 \times \frac{2}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$$ 5. **Find side $x = BC$ (adjacent to $\angle B$):** Using cosine, $$\cos(45^\circ) = \frac{BC}{AB} = \frac{x}{k}$$ So, $$x = k \cos(45^\circ) = 6\sqrt{2} \times \frac{\sqrt{2}}{2} = 6 \times 1 = 6$$ 6. **Triangle ACD:** Given $\angle C = 30^\circ$, side $AD = m$, and side $CD = y$. Since $\angle C = 30^\circ$ and $\angle D = 60^\circ$ (because triangle ACD is right angled at $D$), and $AC = 6$ is the hypotenuse of triangle ACD. 7. **Find side $m = AD$ (opposite $30^\circ$):** $$\sin(30^\circ) = \frac{AD}{AC} = \frac{m}{6}$$ So, $$m = 6 \times \sin(30^\circ) = 6 \times \frac{1}{2} = 3$$ 8. **Find side $y = CD$ (adjacent to $30^\circ$):** $$\cos(30^\circ) = \frac{CD}{AC} = \frac{y}{6}$$ So, $$y = 6 \times \cos(30^\circ) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$$ **Final answers:** $$k = 6\sqrt{2}, \quad m = 3, \quad x = 6, \quad y = 3\sqrt{3}$$