1. **State the problem:** We need to find the length of side $HI$ in a right triangle $IJH$ where the right angle is at $J$, side $IJ = \sqrt{22}$, and angle $I = 61^\circ$. 2. **Identify the sides relative to angle $I$:** - $IJ$ is adjacent to angle $I$ (since $J$ is the right angle). - $HI$ is the hypotenuse (opposite the right angle). 3. **Use the cosine function:** $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ Here, $\theta = 61^\circ$, adjacent side $= IJ = \sqrt{22}$, hypotenuse $= HI$. 4. **Set up the equation:** $$\cos(61^\circ) = \frac{\sqrt{22}}{HI}$$ 5. **Solve for $HI$:** $$HI = \frac{\sqrt{22}}{\cos(61^\circ)}$$ 6. **Calculate the value:** - $\sqrt{22} \approx 4.6904$ - $\cos(61^\circ) \approx 0.4848$ $$HI \approx \frac{4.6904}{0.4848} \approx 9.67$$ **Final answer:** $$HI \approx 9.67$$