Are they equal

Both triangles lie on two congruent circles. By the inscribed angle theorem the angles are equal.

We can express the blue angle as $\beta=\tan^{-1}\left(\dfrac{1}{3}\right)$

And the red angle $\alpha$ as the difference between $\alpha_1=\tan^{-1}1$ and $\alpha_2=\tan^{-1}\left(\dfrac{1}{2}\right)$

Since $\tan(x)$ is strictly increasing in the interval $[0,2\pi)$ , we can compare $\tan\alpha$ and $\tan\beta$ instead

$\tan\alpha=\tan(\alpha_1-\alpha_2)=\dfrac{\tan\alpha_1-\tan\alpha_2}{1+\tan\alpha_1\cdot\tan\alpha_2}=\dfrac{1-\frac{1}{2}}{1+\frac{1}{2}}=\dfrac{1}{3}=\tan\beta$

Since $\tan(x)$ is injective in the interval $[0,2\pi)$ , this implies $\boxed{\alpha=\beta}$