1. **Problem statement:** We have a right triangle ABC with a right angle at B and a smaller triangle ABD inside it. Given $AC = 19$ cm, and angles $\angle B = 46^\circ$ and $\angle D = 35^\circ$ in triangle ABD, find the length $AD$. 2. **Identify the triangle and angles:** In triangle ABD, the angles are $\angle B = 46^\circ$, $\angle D = 35^\circ$, and $\angle A = 180^\circ - 46^\circ - 35^\circ = 99^\circ$. 3. **Use the Law of Sines:** The Law of Sines states: $$\frac{AD}{\sin 46^\circ} = \frac{AB}{\sin 35^\circ} = \frac{BD}{\sin 99^\circ}$$ 4. **Relate $AD$ and $AC$:** Since $AC = 19$ cm and $AD$ is part of $AC$, we need to express $AD$ in terms of known quantities. Note that $AC = AD + DC$, but $DC$ is not given. However, since $B$ is the right angle in triangle ABC, and $D$ lies on $AC$, we can use the angles to find $AD$. 5. **Calculate $AD$ using the Law of Sines in triangle ABD:** We know $AC = 19$ cm, and $AC$ corresponds to side $BD$ in triangle ABD (since $B$ and $D$ are points on the triangle). But this is ambiguous without more info. Instead, consider triangle ABD with angles $46^\circ$, $35^\circ$, and $99^\circ$ and side $AC = 19$ cm as the base. 6. **Assuming $AC$ corresponds to side $AB + BD$ and $AD$ is the side opposite $\angle B$:** Using the Law of Sines: $$\frac{AD}{\sin 46^\circ} = \frac{AC}{\sin 35^\circ}$$ 7. **Solve for $AD$:** $$AD = AC \times \frac{\sin 46^\circ}{\sin 35^\circ} = 19 \times \frac{\sin 46^\circ}{\sin 35^\circ}$$ 8. **Calculate the sine values:** $$\sin 46^\circ \approx 0.7193, \quad \sin 35^\circ \approx 0.5740$$ 9. **Calculate $AD$:** $$AD = 19 \times \frac{0.7193}{0.5740} \approx 19 \times 1.2529 = 23.8$$ 10. **Final answer:** $$\boxed{AD = 23.8 \text{ cm}}$$ (correct to 1 decimal place)