1. **State the problem:** We are given two angles at point H: $m\angle 2 = (\alpha + 15^\circ)$ and $m\angle 3 = (\alpha + 35^\circ)$. We need to find the value of $\alpha$ such that ray $HL$ is perpendicular to ray $JI$. 2. **Analyze the figure and angles:** From the description, rays $HG$, $HJ$, $HK$, $HL$, and $HI$ are arranged around point $H$. Specifically, $HL$ is the ray we want perpendicular to $JI$. Since $JI$ is along the horizontal line $GHI$, and $HL$ is upward-right, for $HL \perp JI$, the angle between $HL$ and $HI$ must be $90^\circ$. 3. **Identify relevant angles:** Angles 2 and 3 are adjacent angles between rays $HJ$, $HK$, and $HL$. The sum of angles 2 and 3 plus the angle between $HL$ and $HI$ must be $90^\circ$ because $HL$ is perpendicular to $HI$. 4. **Set up the equation:** Since $HL \perp JI$ and $JI$ is horizontal, the sum of $m\angle 2$ and $m\angle 3$ must be $90^\circ$ (the right angle at $H$ between $HL$ and $HI$). So, $$ (\alpha + 15^\circ) + (\alpha + 35^\circ) = 90^\circ $$ 5. **Simplify and solve for $\alpha$:** $$ 2\alpha + 50^\circ = 90^\circ $$ $$ 2\alpha = 90^\circ - 50^\circ $$ $$ 2\alpha = 40^\circ $$ $$ \alpha = \frac{40^\circ}{2} = 20^\circ $$ 6. **Final answer:** The value of $\alpha$ that makes $HL \perp JI$ is $$\boxed{20^\circ}$$