$\def\QQ{\mathbb{Q}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$This answer shows that no such triangle exists, under the assumption that Schanuel's conjecture is true. Schanuel's conjecture says: Let $z_1$, $z_2$, ..., $z_n$ be complex numbers which are linearly independent over $\QQ$. Let $K = \QQ(z_1, z_2, \ldots, z_n, e^{z_1}, e^{z_2}, \ldots, e^{z_n})$. Then the transcendence degree of $K$ over $\QQ$ is at least $n$.
We will need successively stronger theorems about transcendence as we go on; starting with the transcendence of $\pi$, then Lindemann's theorem (if $\alpha \neq 0$ is algebraic, then $e^{\alpha}$ is transcendental) and finally getting to the full Schanuel's conjecture.
Let $\alpha_1$, $\alpha_2$, $\alpha_3$ be the angles you called $\alpha$, $\beta$ and $\theta$, and put $z_j = i \alpha_j$, so
$$K = \QQ(i \alpha_1, i \alpha_2, i \alpha_3, e^{i \alpha_1}, e^{i \alpha_2}, e^{i \alpha_3}).$$
The hypothesis is that we have rational numbers $c_j$ such that
$$\alpha_j - \sin \alpha_j = 2 \pi c_j \qquad (1)$$
and thus
$$i \alpha_j = 2 \pi i c_j + \frac{e^{i \alpha_j} - e^{-i\alpha_j}}{2}. \qquad (2).$$
We also have
$$\alpha_1+\alpha_2 + \alpha_3 = 2 \pi \qquad (3)$$
and therefore
$$e^{i \alpha_1} e^{i \alpha_2} e^{i \alpha_3} = 1 \qquad (4).$$
Lemma 1 None of the $\alpha_j$ can be a rational multiple of $\pi$, except possibly if $\alpha_j = \pi$.
Proof Suppose that $\alpha_j$ is a rational multiple of $\pi$. Then $\sin(\alpha)$ is an algebraic number, and is also a rational multiple of $\pi$. Since $\pi$ is transcendental, the only way for this to happen is if $\sin(\alpha) = 0$, meaning that $\alpha = \pi$. $\square$.
We now break into cases, according to the dimension of the $\QQ$ vector space spanned by the $\alpha_i$.
Case 1: The $\alpha_i$ span a $1$-dimensional $\QQ$ vector space. Since $\alpha_1+\alpha_2+ \alpha_3 = 2 \pi$, this means that all the $\alpha_i$ are in $\QQ \pi$. But, by Lemma 1, the only way that this can happen is if $\alpha_1 = \alpha_2 = \alpha_3 = \pi$, and that doesn't obey $\alpha_1+\alpha_2+\alpha_3 = \pi$.
Case 2: The $\alpha_j$ span a two-dimensional $\QQ$-vector space.
Case 2a: One of the $\alpha_i$, without loss of generality $\alpha_3$, is a rational multiple of $\pi$. By Lemma 1, it must be $\pi$. So we can write $(\alpha_1, \alpha_2, \alpha_3) = (\alpha, \pi - \alpha, \pi)$. Then $\alpha - \sin(\alpha) \in \QQ \pi$ and $(\pi - \alpha) - \sin(\pi - \alpha) \in \QQ \pi$ for some rational numbers $u$ and $v$. Subtracting these formulas from each other, and using that $\sin(\pi - \alpha) = \sin(\alpha)$, we deduce that $\alpha \in \QQ \pi$. But then we violate Lemma 1 again. $\square$
Case 2b: None of the $\alpha_j$ is a rational multiple of $\pi$.
By $(3)$, the $\QQ$-vector space spanned by the $\alpha_j$ contains $2 \pi$. Let $(2 \pi, \delta)$ be a basis of this vector space and write
$$\alpha_j = u_j (2\pi) + v_j \delta. \qquad (5)$$
Rescaling $\delta$, we may assume that all the $v_j$ are integers.
By our assumption that $\alpha_j \not\in \QQ \pi$, the $v_j$ are not zero. Also, by $(3)$, we have $u_1+u_2+u_3 = 1$ and $v_1+v_2+v_3=0$.
Since $v_1+v_2+v_3 = 0$, note that it is not possible that $|v_1| = |v_2| = |v_3|$.
Put $\eta = e^{i \delta}$ and put $\zeta_j = e^{(2 \pi i) u_j}$, so
$$e^{i \alpha_j} = \zeta_j \eta^{v_j}. \qquad (6)$$
Note that the $\zeta_j$ are roots of unity, and hence algebraic. Combining $(2)$, $(5)$ and $(6)$, we get
$$\zeta_j^{-1} \eta^{-v_j} - \zeta_j \eta^{v_j} =
(u_j - c_j) (2 \pi i) + v_j (\delta i) \qquad (7).$$
Now, two linear equations in three variables always have a nonzero solution, so we can find nonzero rational constants $(p_1, p_2, p_3)$ such that $\sum p_j (u_j-c_j) = \sum p_j v_j = 0$, and we then have
$$ \sum_j p_j \left( \zeta_j^{-1} \eta^{-v_j} - \zeta_j \eta^{v_j} \right) = 0. \qquad (8)$$
Remembering that the $v_j$ are integers, $(8)$ is a Laurent polynomial equation satisfied by $\eta$, with algebraic coefficients. Remembering that the $v_j$ are not zero, and remembering that we do not have $|v_1| = |v_2| = |v_3|$, it is not the zero polynomial. So $\eta$ is algebraic.
Now, if the matrix $\left[ \begin{smallmatrix} u_1 - c_1 & v_1 \\ u_2 - c_2 & v_2 \\ u_3 - c_3 & v_3 \\ \end{smallmatrix} \right]$ has rank $2$, then I can finish just using that $\pi$ is transcendental. In that case, we can solve $(7)$ as a system of (redundant) linear equations for $\pi$ and $\delta$ to obtain $\pi$ as a rational linear combination of powers of $\eta$. But this contradicts that $\pi$ is transcendental and $\eta$ is algebraic.
Without an assumption on the rank of the matrix, I can still reach a contradiction, but I need the harder Lindemann's theorem. Define
$$\beta_j := \zeta_j^{-1} \eta^{-v_j} - \zeta_j \eta^{v_j} =
(u_j - c_j) (2 \pi i) + v_j (\delta i).$$
(See equation $(7)$.)
Since $\zeta_j$ and $\eta$ are algebraic, $\beta_j$ is algebraic. Since $\pi$ and $\delta$ are linearly independent over $\QQ$ and $v_j$ is nonzero, $\beta_j$ is nonzero. So, by Lindemann's theorem, $e^{\beta_j}$ is transcendental. But, using the right hand side, $e^{\beta_j}$ is the product of a root of unity and $\eta^{v_j}$, so $e^{\beta_j}$ is algebraic; a contradiction.
Case 3: The $\alpha_j$ span a three dimensional $\QQ$-vector space. Now we will need Schanuel's conjecture.
Lemma 2: $\sin \alpha_1 + \sin \alpha_2 + \sin \alpha_3 \neq 0$.
Proof If our triangle is acute, then $0 < \alpha_1, \alpha_2, \alpha_3 < \pi$, so all the sines are positive and the result is clear. For obtuse triangles, we need to work slightly harder: Without loss of generality, let $\alpha_3 \geq \pi$ and $\alpha_1$, $\alpha_2 < \pi$. We have $\sin(\alpha_3) = - \sin(\alpha_1+\alpha_2)$, so we want to rule out the possibility that $\sin(\alpha_1+\alpha_2) = \sin(\alpha_1)+\sin(\alpha_2)$. This is impossible because $\sin$ is concave down on $[0,\pi]$. $\square$
Rewrite equation (1) as
$$\sin(\alpha_j) = \alpha_j - c_j (2 \pi) = \alpha_j - c_j (\alpha_1+\alpha_2+\alpha_3) \qquad (9)$$
and then rewrite this equation in matrix form as
$$\begin{bmatrix} \sin(\alpha_1) \\ \sin(\alpha_2) \\ \sin(\alpha_3) \\ \end{bmatrix} =
\left( \text{Id}_3 - \begin{bmatrix} c_1&c_1&c_1 \\ c_2&c_2&c_2 \\ c_3&c_3&c_3 \end{bmatrix} \right) \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \end{bmatrix}. \qquad (10)$$
We want the matrix on the right hand side of $(10)$ to be invertible, in other words, we want $1$ not to be an eigenvalue of $\left[ \begin{smallmatrix} c_1&c_1&c_1 \\ c_2&c_2&c_2 \\ c_3&c_3&c_3 \end{smallmatrix} \right]$. The eigenvalues of $\left[ \begin{smallmatrix} c_1&c_1&c_1 \\ c_2&c_2&c_2 \\ c_3&c_3&c_3 \end{smallmatrix} \right]$ are $0$, $0$ and $c_1+c_2+c_3$, so we want to rule out the possibility that $c_1+c_2+c_3=1$. If $c_1+c_2+c_3=1$, then adding up the three copies of $(9)$ gives $\sin \alpha_1 + \sin \alpha_2 + \sin \alpha_3=0$, contradicting Lemma 2.
So we have deduced that the matrix in $(10)$ is invertible, and we can write
$$\begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \end{bmatrix} =
\left( \text{Id}_3 - \begin{bmatrix} c_1&c_1&c_1 \\ c_2&c_2&c_2 \\ c_3&c_3&c_3 \end{bmatrix} \right)^{-1}
\begin{bmatrix} \sin(\alpha_1) \\ \sin(\alpha_2) \\ \sin(\alpha_3) \\ \end{bmatrix}$$
and thus
$$\begin{bmatrix} i \alpha_1 \\ i \alpha_2 \\ i \alpha_3 \\ \end{bmatrix} =
\left( \text{Id}_3 - \begin{bmatrix} c_1&c_1&c_1 \\ c_2&c_2&c_2 \\ c_3&c_3&c_3 \end{bmatrix} \right)^{-1}
\begin{bmatrix} (e^{-i\alpha_1} - e^{i \alpha_1})/2 \\ (e^{-i\alpha_1} - e^{i \alpha_1})/2 \\ (e^{-i\alpha_1} - e^{i \alpha_1})/2 \\ \end{bmatrix}.$$
So
$$\QQ(i \alpha_1, i \alpha_2, i \alpha_3, e^{i \alpha_1}, e^{i \alpha_2}, e^{i \alpha_3}) = \QQ(e^{i \alpha_1}, e^{i \alpha_2}, e^{i \alpha_3}).$$
But then the relation $(4)$ shows that $\QQ(e^{i \alpha_1}, e^{i \alpha_2}, e^{i \alpha_3})=\QQ(e^{i \alpha_1}, e^{i \alpha_2})$, so the transcendence degree of $K$ over $\QQ$ is at most $2$, violating Schanuel's conjecture.
