1. **Problem statement:** We have a cylindrical roller of diameter 25 units resting in a V-shaped groove with two sides inclined at 122° each. The horizontal distance between the groove sides is $l=38$ units nominally. We need to find the vertical height $h$ from the base to a horizontal line touching the roller. 2. **Given:** - Diameter of roller $D=25$ units - Radius $r=\frac{D}{2}=12.5$ units - Angle of each groove side $\theta=122^\circ$ - Horizontal distance $l=38$ units nominal 3. **Step 1: Find $h$ using geometry.** The groove forms a V with two sides inclined at $122^\circ$ to the horizontal, so the angle between the two sides is $2\times(180^\circ-122^\circ)=116^\circ$. The roller touches both sides, so its center lies on the bisector of the angle between the sides. The horizontal distance $l$ is the distance between the two contact points on the sides. Using the geometry of the roller in the groove, the vertical height $h$ from the base to the center of the roller is: $$h = r \times \cot\left(\frac{116^\circ}{2}\right) = 12.5 \times \cot(58^\circ)$$ Calculate $\cot(58^\circ)$: $$\cot(58^\circ) = \frac{1}{\tan(58^\circ)} \approx \frac{1}{1.6003} = 0.6249$$ So, $$h = 12.5 \times 0.6249 = 7.81125$$ Rounded to nearest 0.002 units: $$h = 7.812$$ 4. **Step 2: Determine error in $l$ when each 122° angle is oversize by 4 minutes of arc.** 4 minutes of arc = $\frac{4}{60} = 0.0667^\circ$ New angle $\theta' = 122^\circ + 0.0667^\circ = 122.0667^\circ$ New angle between sides: $$2 \times (180^\circ - 122.0667^\circ) = 2 \times 57.9333^\circ = 115.8666^\circ$$ Calculate new $l'$ using the relation: $$l = 2r \sin\left(\frac{\alpha}{2}\right)$$ where $\alpha$ is the angle between the sides. Original $l$: $$l = 2 \times 12.5 \times \sin\left(\frac{116^\circ}{2}\right) = 25 \times \sin(58^\circ)$$ Calculate $\sin(58^\circ)$: $$\sin(58^\circ) \approx 0.8480$$ So original $l$: $$l = 25 \times 0.8480 = 21.2$$ New $l'$: $$l' = 25 \times \sin\left(\frac{115.8666^\circ}{2}\right) = 25 \times \sin(57.9333^\circ)$$ Calculate $\sin(57.9333^\circ)$: $$\sin(57.9333^\circ) \approx 0.8471$$ So, $$l' = 25 \times 0.8471 = 21.1775$$ Error in $l$: $$\Delta l = l' - l = 21.1775 - 21.2 = -0.0225$$ Rounded to nearest 0.002 units: $$\Delta l = -0.022$$ **Interpretation:** The horizontal distance $l$ decreases by approximately 0.022 units due to the oversize angle. --- **Final answers:** $$h = 7.812$$ units $$\text{Error in } l = -0.022$$ units