1. **State the problem:** We have a right triangle BCD with right angle at C. The side opposite angle B is $21\sqrt{6}$, the side opposite angle D is $21\sqrt{2}$, and we need to find the hypotenuse $BD$ and the measures of angles $D$ and $B$. 2. **Use the Pythagorean theorem:** For a right triangle, the hypotenuse squared equals the sum of the squares of the other two sides: $$BD^2 = BC^2 + CD^2$$ 3. **Calculate $BD$:** $$BD^2 = (21\sqrt{6})^2 + (21\sqrt{2})^2$$ $$= 21^2 \times 6 + 21^2 \times 2$$ $$= 441 \times 6 + 441 \times 2$$ $$= 2646 + 882$$ $$= 3528$$ 4. **Simplify $BD$:** $$BD = \sqrt{3528}$$ Factor 3528: $$3528 = 4 \times 882 = 4 \times 9 \times 98 = 4 \times 9 \times 49 \times 2$$ $$BD = \sqrt{4 \times 9 \times 49 \times 2} = \sqrt{4} \times \sqrt{9} \times \sqrt{49} \times \sqrt{2} = 2 \times 3 \times 7 \times \sqrt{2} = 42\sqrt{2}$$ 5. **Find angles $D$ and $B$ using trigonometry:** Use tangent, for example: $$\tan(\angle D) = \frac{\text{opposite side to } D}{\text{adjacent side to } D} = \frac{BC}{CD} = \frac{21\sqrt{6}}{21\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$$ 6. **Calculate $\angle D$:** $$\angle D = \tan^{-1}(\sqrt{3}) = 60^\circ$$ 7. **Calculate $\angle B$:** Since the triangle is right angled at $C$, the sum of angles $B$ and $D$ is $30^\circ$ because: $$\angle B = 90^\circ - \angle D = 90^\circ - 60^\circ = 30^\circ$$ **Final answers:** $$BD = 42\sqrt{2}$$ $$m\angle D = 60^\circ$$ $$m\angle B = 30^\circ$$