Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be unit vectors such that
\[\mathbf{a} + \mathbf{b} + \sqrt{3} \mathbf{c} = \mathbf{0}.\]Find the angle between $\mathbf{a}$ and $\mathbf{b},$ in degrees.

Note: A unit vector is a vector of magnitude 1.
From the given equation,
\[\mathbf{a} + \mathbf{b} = -\sqrt{3} \mathbf{c}.\]Then $(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = 3 \mathbf{c} \cdot \mathbf{c} = 3.$  Expanding, we get
\[\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} = 3.\]Then $2 \mathbf{a} \cdot \mathbf{b} = 1,$ so $\mathbf{a} \cdot \mathbf{b} = \frac{1}{2}.$

If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b},$ then
\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{1/2}{1 \cdot 1} = \frac{1}{2},\]so $\theta = \boxed{60^\circ}.$
60
Level 2
Precalculus
Problem: Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be unit vectors such that
\[\mathbf{a} + \mathbf{b} + \sqrt{3} \mathbf{c} = \mathbf{0}.\]Find the angle between $\mathbf{a}$ and $\mathbf{b},$ in degrees.

Note: A unit vector is a vector of magnitude 1.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$, we start from the equation $\mathbf{a} + \mathbf{b} + \sqrt{3} \mathbf{c} = \mathbf{0}$. Rearranging gives us $\mathbf{a} + \mathbf{b} = -\sqrt{3} \mathbf{c}$. 

Since $\mathbf{c}$ is a unit vector, we have $|\mathbf{c}| = 1$. Therefore, the magnitude of the right-hand side is $|\mathbf{a} + \mathbf{b}| = \sqrt{3}$. 

Using the property of magnitudes, we can express $|\mathbf{a} + \mathbf{b}|$ in terms of the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$:
\[
|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2 |\mathbf{a}| |\mathbf{b}| \cos(\theta).
\]
Since both $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, we have $|\mathbf{a}|^2 = 1$ and $|\mathbf{b}|^2 = 1$. Thus, the equation simplifies to:
\[
|\mathbf{a} + \mathbf{b}|^2 = 1 + 1 + 2 \cos(\theta) = 2 + 2 \cos(\theta).
\]
Setting this equal to the square of the magnitude we found earlier, we have:
\[
2 + 2 \cos(\theta) = 3.
\]
Subtracting 2 from both sides gives:
\[
2 \cos(\theta) = 1.
\]
Dividing by 2 results in:
\[
\cos(\theta) = \frac{1}{2}.
\]
The angle $\theta$ that satisfies this equation is $60$ degrees, as $\cos(60^\circ) = \frac{1}{2}$. Thus, the angle between $\mathbf{a}$ and $\mathbf{b}$ is $60$ degrees.<|END_OF_LATENT|> The final answer is $60$.