1. We are given a right triangle with sides 3, 4, and 5, where the hypotenuse (length 5) is divided into three segments of lengths 1, 2, and 2. 2. The problem asks to find the angle $\alpha$ near the vertex where the sides 3 and 4 meet. 3. Since the triangle is right-angled with sides 3, 4, and 5, the angle $\alpha$ at the vertex between sides 3 and 4 is the right angle, so $\alpha = 90^\circ$. 4. However, the problem suggests $\alpha$ is an angle formed by the division of the hypotenuse into segments 1, 2, and 2, so we need to analyze the geometry more carefully. 5. Let's place the triangle in the coordinate plane: let the right angle be at the origin $O(0,0)$, side of length 4 along the x-axis to point $B(4,0)$, and side of length 3 along the y-axis to point $A(0,3)$. 6. The hypotenuse $AB$ has length 5, and points dividing $AB$ into segments 1, 2, and 2 lie on $AB$. 7. The vector $\overrightarrow{AB} = (4 - 0, 0 - 3) = (4, -3)$. 8. The total length of $AB$ is 5, so the unit vector along $AB$ is $\frac{1}{5}(4, -3) = (\frac{4}{5}, -\frac{3}{5})$. 9. The points dividing $AB$ are: - $P_1$ at distance 1 from $A$: $A + 1 \times$ unit vector $= (0 + \frac{4}{5}, 3 - \frac{3}{5}) = (0.8, 2.4)$ - $P_2$ at distance 3 from $A$ (1 + 2): $A + 3 \times$ unit vector $= (0 + \frac{12}{5}, 3 - \frac{9}{5}) = (2.4, 1.2)$ 10. The angle $\alpha$ is at $O(0,0)$ between side $OA$ (vertical) and the segment $OP_1$ or $OP_2$? The problem states $\alpha$ is near the vertex where sides 3 and 4 meet, so it's angle $AOB$. 11. Let's find the angle between vectors $OA = (0,3)$ and $OB = (4,0)$: $$\cos \alpha = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|OA||OB|} = \frac{0 \times 4 + 3 \times 0}{3 \times 4} = 0$$ 12. So $\alpha = 90^\circ$. 13. But the problem's options do not include 90°, so $\alpha$ must be the angle between the segment dividing the hypotenuse and one of the sides. 14. Let's find the angle between $OP_1$ and $OA$: - Vector $OP_1 = (0.8, 2.4)$ - Vector $OA = (0,3)$ $$\cos \alpha = \frac{0.8 \times 0 + 2.4 \times 3}{\sqrt{0.8^2 + 2.4^2} \times 3} = \frac{7.2}{\sqrt{0.64 + 5.76} \times 3} = \frac{7.2}{\sqrt{6.4} \times 3}$$ $$\sqrt{6.4} \approx 2.5298$$ $$\cos \alpha = \frac{7.2}{2.5298 \times 3} = \frac{7.2}{7.5894} \approx 0.9487$$ 15. Therefore, $$\alpha = \arccos(0.9487) \approx 18.43^\circ$$ 16. This is less than 30°, so not matching options. 17. Now check angle between $OP_2 = (2.4, 1.2)$ and $OA = (0,3)$: $$\cos \alpha = \frac{2.4 \times 0 + 1.2 \times 3}{\sqrt{2.4^2 + 1.2^2} \times 3} = \frac{3.6}{\sqrt{5.76 + 1.44} \times 3} = \frac{3.6}{\sqrt{7.2} \times 3}$$ $$\sqrt{7.2} \approx 2.6833$$ $$\cos \alpha = \frac{3.6}{2.6833 \times 3} = \frac{3.6}{8.0499} \approx 0.4472$$ 18. So, $$\alpha = \arccos(0.4472) \approx 63.43^\circ$$ 19. This is between 60° and 65°, so closest to option (E) $\alpha = 60^\circ$. 20. Since the problem's options are discrete, the best match is (E) $\alpha = 60^\circ$. Final answer: (E) $\alpha = 60^\circ$