Given that $\theta$ and $\phi$ are acute angles such that $\tan \theta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}},$ find $\theta + 2 \phi,$ measured in radians.
Note that
\[\cos^2 \phi = 1 - \sin^2 \phi = \frac{9}{10}.\]Since $\phi$ is acute, $\cos \phi = \frac{3}{\sqrt{10}}.$  Then
\[\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{1}{3},\]so
\[\tan 2 \phi = \frac{2 \tan \phi}{1 - \tan^2 \phi} = \frac{2 \cdot \frac{1}{3}}{1 - (\frac{1}{3})^2} = \frac{3}{4},\]and
\[\tan (\theta + 2 \phi) = \frac{\tan \theta + \tan 2 \phi}{1 - \tan \theta \tan 2 \phi} = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \cdot \frac{3}{4}} = 1.\]Since $\tan 2 \phi$ is positive, $2 \phi$ is also acute.  Hence, $0 < \theta + 2 \phi < \pi.$   Therefore, $\theta + 2 \phi = \boxed{\frac{\pi}{4}}.$
\frac{\pi}{4}
Level 4
Precalculus
Problem: Given that $\theta$ and $\phi$ are acute angles such that $\tan \theta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}},$ find $\theta + 2 \phi,$ measured in radians.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the angles $\theta$ and $\phi$, we start with the given values. For $\tan \theta = \frac{1}{7}$, we can use the definition of tangent in terms of sine and cosine. This implies that $\sin \theta = \frac{1}{\sqrt{1^2 + 7^2}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}}$ and $\cos \theta = \frac{7}{\sqrt{50}} = \frac{7}{5\sqrt{2}}$. 

Next, we consider $\sin \phi = \frac{1}{\sqrt{10}}$. To find $\cos \phi$, we use the Pythagorean identity $\cos^2 \phi = 1 - \sin^2 \phi$. Thus, $\cos^2 \phi = 1 - \left(\frac{1}{\sqrt{10}}\right)^2 = 1 - \frac{1}{10} = \frac{9}{10}$, leading to $\cos \phi = \frac{3}{\sqrt{10}}$.

Now, we need to compute $\theta + 2\phi$. We can find $\tan 2\phi$ using the double angle formula: $\tan 2\phi = \frac{2\tan \phi}{1 - \tan^2 \phi}$. First, we find $\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}$. 

Now, substituting into the double angle formula gives us $\tan 2\phi = \frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2 \cdot 9}{3 \cdot 8} = \frac{3}{4}$. 

Next, we find $\tan(\theta + 2\phi)$ using the tangent addition formula: $\tan(\theta + 2\phi) = \frac{\tan \theta + \tan 2\phi}{1 - \tan \theta \tan 2\phi}$. Substituting $\tan \theta = \frac{1}{7}$ and $\tan 2\phi = \frac{3}{4}$ gives us $\tan(\theta + 2\phi) = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \cdot \frac{3}{4}}$. 

Calculating the numerator: $\frac{1}{7} + \frac{3}{4} = \frac{4 + 21}{28} = \frac{25}{28}$. For the denominator: $1 - \frac{3}{28} = \frac{25}{28}$. Thus, $\tan(\theta + 2\phi) = \frac{\frac{25}{28}}{\frac{25}{28}} = 1$. 

Since $\tan(\theta + 2\phi) = 1$, we conclude that $\theta + 2\phi = \frac{\pi}{4}$, as the tangent of $\frac{\pi}{4}$ is 1. Therefore, the final answer is $\frac{\pi}{4}$.<|END_OF_LATENT|> The final answer is $\frac{\pi}{4}$.