1. **State the problem:** We are given a right triangle MEJ (labeled KIJ in the diagram) with a right angle at vertex I. Given sides: - KI = $4\sqrt{6}$ - IJ = $4\sqrt{2}$ We need to find: - Hypotenuse JK - Angle $\angle K$ - Angle $\angle J$ 2. **Use the Pythagorean theorem:** For a right triangle with legs $a$ and $b$, and hypotenuse $c$, the relation is: $$c^2 = a^2 + b^2$$ Here, $a = KI = 4\sqrt{6}$ and $b = IJ = 4\sqrt{2}$. 3. **Calculate the hypotenuse JK:** $$JK^2 = (4\sqrt{6})^2 + (4\sqrt{2})^2$$ $$= 16 \times 6 + 16 \times 2$$ $$= 96 + 32 = 128$$ So, $$JK = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$$ 4. **Find the angles using trigonometry:** Since $\angle I$ is the right angle, angles $\angle K$ and $\angle J$ are acute. Use tangent to find $\angle K$: $$\tan(\angle K) = \frac{\text{opposite}}{\text{adjacent}} = \frac{IJ}{KI} = \frac{4\sqrt{2}}{4\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{6}}$$ Simplify: $$\frac{\sqrt{2}}{\sqrt{6}} = \sqrt{\frac{2}{6}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$$ So, $$\tan(\angle K) = \frac{1}{\sqrt{3}}$$ From special angles, $$\tan(30^\circ) = \frac{1}{\sqrt{3}}$$ Therefore, $$\angle K = 30^\circ$$ 5. **Find $\angle J$:** Since the sum of angles in a triangle is $180^\circ$ and $\angle I = 90^\circ$, $$\angle J = 180^\circ - 90^\circ - 30^\circ = 60^\circ$$ **Final answers:** - $JK = 8\sqrt{2}$ - $m\angle K = 30^\circ$ - $m\angle J = 60^\circ$