Let $\mathbf{a}$ and $\mathbf{b}$ be nonzero vectors such that
\[\|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{a} + \mathbf{b}\|.\]Find the angle between $\mathbf{a}$ and $\mathbf{b},$ in degrees.
Let $d = \|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{a} + \mathbf{b}\|.$  Then
\begin{align*}
d^2 &= \|\mathbf{a} + \mathbf{b}\|^2 \\
&= (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \\
&= \|\mathbf{a}\|^2 + 2 \mathbf{a} \cdot \mathbf{b} + \|\mathbf{b}\|^2 \\
&= 2d^2 + 2 \mathbf{a} \cdot \mathbf{b},
\end{align*}so $\mathbf{a} \cdot \mathbf{b} = -\frac{d^2}{2}.$

Hence, if $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b},$ then
\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{-\frac{d^2}{2}}{d^2} = -\frac{1}{2},\]so $\theta = \boxed{120^\circ}.$
120
Level 5
Precalculus
Problem: Let $\mathbf{a}$ and $\mathbf{b}$ be nonzero vectors such that
\[\|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{a} + \mathbf{b}\|.\]Find the angle between $\mathbf{a}$ and $\mathbf{b},$ in degrees.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$, we start with the given condition that the magnitudes of the vectors are equal and also equal to the magnitude of their sum. This can be expressed mathematically as:

1. Let $\|\mathbf{a}\| = \|\mathbf{b\|} = k$ for some positive scalar $k$.
2. The magnitude of the sum of the vectors can be expressed using the cosine of the angle $\theta$ between them: 
   \[
   \|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + 2\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta).
   \]
3. Substituting the known magnitudes into the equation gives:
   \[
   \|\mathbf{a} + \mathbf{b}\|^2 = k^2 + k^2 + 2k^2\cos(\theta) = 2k^2(1 + \cos(\theta)).
   \]
4. Since we also know that $\|\mathbf{a} + \mathbf{b}\| = k$, we can square both sides:
   \[
   \|\mathbf{a} + \mathbf{b}\|^2 = k^2.
   \]
5. Setting the two expressions for $\|\mathbf{a} + \mathbf{b}\|^2$ equal to each other gives:
   \[
   2k^2(1 + \cos(\theta)) = k^2.
   \]
6. Dividing both sides by $k^2$ (which is nonzero) simplifies to:
   \[
   2(1 + \cos(\theta)) = 1.
   \]
7. Rearranging this equation leads to:
   \[
   2 + 2\cos(\theta) = 1 \implies 2\cos(\theta) = -1 \implies \cos(\theta) = -\frac{1}{2}.
   \]
8. The angle $\theta$ that satisfies $\cos(\theta) = -\frac{1}{2}$ is $120$ degrees, as this corresponds to the second quadrant where cosine is negative.

Thus, the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is $120$ degrees.<|END_OF_LATENT|> The final answer is $120$.