1. **State the problem:** We need to find the length of side $CD$ in triangle $ABCD$ where $AB$ and $AD$ are perpendicular, $AD=4.2$ cm, angle $B=47^\circ$, and angle $D=26^\circ$. 2. **Identify known angles and sides:** Since $AB$ and $AD$ are perpendicular, angle $A=90^\circ$. 3. **Calculate angle $C$:** The sum of angles in a triangle is $180^\circ$, so in triangle $BCD$: $$\angle B + \angle C + \angle D = 180^\circ$$ $$47^\circ + \angle C + 26^\circ = 180^\circ$$ $$\angle C = 180^\circ - 47^\circ - 26^\circ = 107^\circ$$ 4. **Use Law of Sines in triangle $ABD$ to find $AB$:** $$\frac{AB}{\sin 26^\circ} = \frac{4.2}{\sin 47^\circ}$$ $$AB = \frac{4.2 \times \sin 26^\circ}{\sin 47^\circ}$$ Calculate intermediate step: $$AB = \frac{4.2 \times 0.4384}{0.7314} = \frac{1.8408}{0.7314}$$ $$AB \approx 2.516$$ 5. **Use Law of Sines in triangle $BCD$ to find $CD$:** $$\frac{CD}{\sin 47^\circ} = \frac{BC}{\sin 107^\circ}$$ We need $BC$ first. Since $BC = AB$ (from the figure, $B$ is common vertex and $AB$ is horizontal side), we take $BC = AB = 2.516$ cm. 6. **Calculate $CD$:** $$CD = \frac{2.516 \times \sin 47^\circ}{\sin 107^\circ}$$ Calculate intermediate step: $$CD = \frac{2.516 \times 0.7314}{0.9563} = \frac{1.8408}{0.9563}$$ $$CD \approx 1.925$$ 7. **Round to nearest tenth:** $$CD \approx 1.9 \text{ cm}$$ **Final answer:** The length of $CD$ is approximately **1.9 cm**.