The example by alex.jordan does finish the matter, and similar ones may be constructed. We have an angle $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $ and we wish to know whether $ x = \frac{\theta}{\pi} $ is the root of an equation with rational coefficients.
Well, $ e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ Next, $\cos 2 \theta = 2 \cos^2 \theta - 1 = \frac{11}{13}.$ So, by Corollary 3.12 on page 41 of NIVEN we know that $2 \theta$ is not a rational multiple of $\pi.$ So, neither is $\theta,$ and $ x = \frac{\theta}{\pi} $ is irrational.
Now, the logarithm is multivalued in the complex plane. We may choose $ \log(-1) = \pi i. $ With real $x,$ we have chosen $ (-1)^x = \exp(x \log(-1)) = \exp(x\pi i) = \cos \pi x + i \sin \pi x. $ With our $ x = \frac{\theta}{\pi}, $ we have $ (-1)^x = e^{i \pi x} = e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ The right hand side is algebraic.
The Gelfond-Schneider Theorem, Niven page 134, says that if $\alpha,\beta$ are nonzero algebraic numbers, with $\alpha \neq 1$ and $\beta$ not a real rational number, then any value of $\alpha^\beta$ is transcendental.
Taking $\alpha = -1$ and $\beta = x,$ which is real but irrational. We are ASSUMING that $x$ is algebraic over $\mathbb Q.$ The assumption, together with Gelfond-Schneider, says that $ (-1)^x$ is transcendental. However, we already know that $ (-1)^x = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ is algebraic. This contradicts the assumption. So $x = \theta / \pi$ is transcendental, with $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $