1. **Problem statement:** Given a right triangle ABC with right angle at B, side BC = 16 cm, and angles 58° and 41° at points D and A respectively, find the length of AD. 2. **Understanding the problem:** We know BC = 16 cm, and angles at D and A are 58° and 41°. Since the triangle is right angled at B, angle ABC = 90°. 3. **Using triangle angle sum:** In triangle BDC, angle BDC = 58°, angle B = 90°, so angle DCB = 180° - 90° - 58° = 32°. 4. **Using triangle angle sum in triangle ADC:** Angle at A is 41°, angle at D is 58°, so angle ADC = 180° - 41° - 58° = 81°. 5. **Using Law of Sines in triangle BDC:** $$\frac{BD}{\sin 32^\circ} = \frac{BC}{\sin 58^\circ}$$ Substitute BC = 16: $$\frac{BD}{\sin 32^\circ} = \frac{16}{\sin 58^\circ}$$ Solve for BD: $$BD = \frac{16 \times \sin 32^\circ}{\sin 58^\circ}$$ Calculate numerically: $$BD \approx \frac{16 \times 0.5299}{0.8480} \approx 10.0\, \text{cm}$$ 6. **Using Law of Sines in triangle ADC:** $$\frac{AD}{\sin 58^\circ} = \frac{BD}{\sin 81^\circ}$$ Solve for AD: $$AD = \frac{BD \times \sin 58^\circ}{\sin 81^\circ}$$ Substitute BD \approx 10.0: $$AD \approx \frac{10.0 \times 0.8480}{0.9877} \approx 8.6\, \text{cm}$$ 7. **Final answer:** $$\boxed{AD = 8.6\, \text{cm}}$$ This is the length of AD correct to 1 decimal place.