If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a}+2\vec{b}$ and $5\vec{a}-4\vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is:

Solution:

Given two unit vectors \( \vec{a} \) and \( \vec{b} \), and the vectors \( \vec{a} + 2\vec{b} \) and \( 5\vec{a} - 4\vec{b} \) are perpendicular, we use the condition for perpendicular vectors: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 $$ Expanding the dot product: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b}) $$ Using properties of dot products and knowing \( \vec{a} \) and \( \vec{b} \) are unit vectors (\( \vec{a} \cdot \vec{a} = 1 \) and \( \vec{b} \cdot \vec{b} = 1 \)): $$ 5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0 $$ Simplifying: $$ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 $$ $$ 5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ -3 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ 6(\vec{a} \cdot \vec{b}) = 3 $$ $$ \vec{a} \cdot \vec{b} = \frac{1}{2} $$ The dot product \( \vec{a} \cdot \vec{b} = \cos \theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \): $$ \cos \theta = \frac{1}{2} $$ Therefore, the angle \( \theta \) is: $$ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ $$
Final Answer:
$$ \boxed{60^\circ} $$