Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors such that
\[\|\mathbf{a} + \mathbf{b}\| = \|\mathbf{b}\|.\]Find the angle between the vectors $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a},$ in degrees
From the equation $\|\mathbf{a} + \mathbf{b}\| = \|\mathbf{b}\|,$ $\|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{b}\|^2,$ so
\[(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{b} \cdot \mathbf{b}.\]Expanding, we get $\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{b},$ so
\[\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} = 0.\]We can write this as $\mathbf{a} \cdot (\mathbf{a} + 2 \mathbf{b}) = 0.$  Thus, the vectors $\mathbf{a}$ and $\mathbf{a} + 2 \mathbf{b}$ are orthogonal, and the angle between them is $\boxed{90^\circ}.$
90
Level 3
Precalculus
Problem: Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors such that
\[\|\mathbf{a} + \mathbf{b}\| = \|\mathbf{b}\|.\]Find the angle between the vectors $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a},$ in degrees
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the angle between the vectors $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a}$, we start by using the given condition $\|\mathbf{a} + \mathbf{b}\| = \|\mathbf{b}\|$. This implies that the vector $\mathbf{a} + \mathbf{b}$ is orthogonal to $\mathbf{b}$, since the only way for the norm of a vector to equal the norm of another vector while being the sum of them is if the first vector is perpendicular to the second.

We can express this condition mathematically. The squared norm gives us:
\[
\|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{b}\|^2.
\]
Expanding the left side, we have:
\[
\|\mathbf{a}\|^2 + 2 \langle \mathbf{a}, \mathbf{b} \rangle + \|\mathbf{b}\|^2 = \|\mathbf{b}\|^2.
\]
Subtracting $\|\mathbf{b}\|^2$ from both sides results in:
\[
\|\mathbf{a}\|^2 + 2 \langle \mathbf{a}, \mathbf{b} \rangle = 0.
\]
This implies that:
\[
\|\mathbf{a}\|^2 = -2 \langle \mathbf{a}, \mathbf{b} \rangle.
\]
Since the left side is non-negative, it follows that $\langle \mathbf{a}, \mathbf{b} \rangle$ must be non-positive, indicating that the angle between $\mathbf{a}$ and $\mathbf{b}$ is either $90$ degrees or obtuse.

Next, we need to find the angle between $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a}$. We can use the dot product to find this angle. The dot product is given by:
\[
\langle \mathbf{a} + 2 \mathbf{b}, \mathbf{a} \rangle = \langle \mathbf{a}, \mathbf{a} \rangle + 2 \langle \mathbf{b}, \mathbf{a} \rangle = \|\mathbf{a}\|^2 + 2 \langle \mathbf{b}, \mathbf{a} \rangle.
\]
Substituting $\langle \mathbf{b}, \mathbf{a} \rangle = -\frac{1}{2} \|\mathbf{a}\|^2$ from our earlier derivation, we get:
\[
\langle \mathbf{a} + 2 \mathbf{b}, \mathbf{a} \rangle = \|\mathbf{a}\|^2 + 2 \left(-\frac{1}{2} \|\mathbf{a}\|^2\right) = \|\mathbf{a}\|^2 - \|\mathbf{a}\|^2 = 0.
\]
Since the dot product is zero, it indicates that the vectors $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a}$ are orthogonal.

Thus, the angle between the vectors $\mathbf{a} + 2 \mathbf{b}$ and $\mathbf{a}$ is $90$ degrees.<|END_OF_LATENT|> The final answer is $90$.