1. **Problem statement:** We have a right triangle with a hypotenuse of length $13\sqrt{2}$, angles $45^\circ$ and $30^\circ$, and a perpendicular dropped from the top vertex to the base, creating segments $a$, $b$, $c$, and $d$. We need to find the exact values of $a$, $b$, $c$, and $d$. 2. **Key facts and formulas:** - The hypotenuse length is $13\sqrt{2}$. - The triangle is split into two right triangles by the perpendicular. - Use trigonometric ratios: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. - The angles are $45^\circ$ and $30^\circ$, so the third angle is $105^\circ$ but the right angle is at the foot of the perpendicular. 3. **Find side $b$ opposite $30^\circ$:** $$b = 13\sqrt{2} \times \sin 30^\circ = 13\sqrt{2} \times \frac{1}{2} = \frac{13\sqrt{2}}{2}$$ 4. **Find side $a$ (the perpendicular height):** Since $a$ is opposite the right angle in the smaller triangle with angle $45^\circ$, use: $$a = 13\sqrt{2} \times \sin 45^\circ = 13\sqrt{2} \times \frac{\sqrt{2}}{2} = 13 \times \cancel{\sqrt{2}} \times \frac{\cancel{\sqrt{2}}}{2} = 13 \times \frac{2}{2} = 13$$ 5. **Find segment $c$ adjacent to $45^\circ$ angle:** $$c = 13\sqrt{2} \times \cos 45^\circ = 13\sqrt{2} \times \frac{\sqrt{2}}{2} = 13$$ 6. **Find segment $d$ adjacent to $30^\circ$ angle:** $$d = 13\sqrt{2} \times \cos 30^\circ = 13\sqrt{2} \times \frac{\sqrt{3}}{2} = \frac{13\sqrt{6}}{2}$$ **Final answers:** $$a = 13, \quad b = \frac{13\sqrt{2}}{2}, \quad c = 13, \quad d = \frac{13\sqrt{6}}{2}$$