It can be verified that for a rational number not nule $q$ we have $\sin{q}$ is irrational.Actually it is transcendent,but we don't need.

mateivld wrote: It can be verified that for a rational number not nule $q$ we have $\sin{q}$ is irrational. Actually it is transcendent,but we don't need. This does not answer the question. Surely $30^\circ,\ 90^\circ,\ 150^\circ \in (0^\circ,\ 180^\circ)$ are all rational (in degrees) and $\sin 30^\circ = \sin 150^\circ = \frac{1}{2},\ \sin 90^\circ = 1$ are also rational. Besides, plenty of angles rational in degrees have sines algebraic irrational. (All of them, except for those few with rational sines.) By the problem condition, the circumradius $R = \frac{abc}{4S}$ is rational, hence $\sin A = \frac{a}{2R},\ \sin B = \frac{b}{2R},\ \sin C = \frac{c}{2R}$ are all rational. We cannot make a triangle with any selection of angles from $\{30^\circ,\ 90^\circ,\ 150^\circ\},$ we need some other angle $\theta$ rational in degrees, such that $\sin \theta$ is rational. Let $\theta \in [0^\circ,\ 180^\circ].$ Then $\phi = 90^\circ-\theta \in [-90^\circ,\ +90^\circ],$ $\phi = \frac{m}{n}\cdot 360^\circ$ (where $n > 0,$ $-n \le 4m \le+n$) is also rational, and $\cos \phi = \sin \theta,\ 2 \cos \phi = \frac{p}{q}$ are also rational. We can take p, q so that $p \ge 0,\ q > 0,\ \gcd(p,\ q) = 1.$ Since $0 \le \cos \phi \le 1,$ we also have $p \le 2q.$ $2 \cos 2 \phi = 4 \cos^{2}\phi-2= \frac{p^{2}}{q^{2}}-2= \frac{p^{2}-2q^{2}}{q^{2}}.$ Assume that $\gcd(p^{2}-2q^{2},\ q^{2}) = d > 1.$ Since d divides $q^{2},$ it divides both q and $2q^{2}$, and since d divides both $p^{2}-2q^{2}$ and $2q^{2},$ it also divides $p^{2}$ and consequently p, so that $\gcd(p,\ q) \ge d > 1,$ which is a contradiction. Hence $\gcd(p^{2}-2q^{2},\ q^{2}) = 1$ as well. Assuming q > 1, the denominators $q_{k}= q^{2^{k}}$ of $2 \cos 2^{k}\phi = \frac{p_{k}}{q_{k}}$ then grow without bounds for $k \longrightarrow \infty.$ But cosine is a periodic function with the period $360^\circ.$ Thus for every n, the sequence $\frac{p_{k}}{q_{k}}= 2 \cos 2^{k}\phi = 2 \cos \left(\frac{2^{k}m}{n}\cdot 360^\circ\right),$ $k = 0,\ 1,\ 2,\ ...$ contains at most n different values $2 \cos \left(\frac{l}{n}\cdot 360^\circ\right),\ l = 0,\ 1,\ ...,\ n-1,$ which contradicts $q_{k}\longrightarrow \infty.$ As a result, q = 1 and the only rational cosines of rational angles $-90^\circ \le \phi \le 90^\circ$ are given by $2 \cos \phi = p \le 2,$ where p = 0, p = 1, or p = 2. Hence $\cos \phi = 0,$ $\cos \phi = \frac{1}{2},$ or $\cos \phi = 1,$ so that $\phi \in \{0^\circ,\ \pm 60^\circ,\ \pm 90^\circ \}$ and $\theta = 90^\circ-\phi \in \{0^\circ,\ 30^\circ,\ 90^\circ,\ 150^\circ,\ 180^\circ \}.$ In conclusion, triangles satisfying the problem condition do not exist.

is it a known problem :

anonymous1173 wrote: is it a known problem : I think that it is connected with the well-known problem of triangles with integer lenghts od heights, angle bisectors and sides.

Yes, it is a well known problem. Another way is to consider complex roots of poynomial $z^{n}-1 = 0,$ because an angle rational in degrees $\theta = \frac{m}{n}\cdot 360^\circ$ is the angle $\angle A_{0}OA_{m}$ of a regular n-gon $A_{0}A_{1}...A_{n-1}$ with center O. This polynomial has roots $z_{m} = \cos \left(\frac{m}{n} \cdot 360^\circ\right) +$ $ \text{i}\sin\left(\frac{m}{n} \cdot 360^\circ\right),$ $m = 0,\ 1,\ ...,\ n-1.$ The polynomial in z can be transformed into a polynomial in $x = 2 \cos \theta$ with integer coefficients and leading coefficient 1, by factoring as much as possible and then substituting $x = z+\frac{1}{z}$ in each factor. By the rational root theorem, the only possible rational roots of the polynomial in x are integers, and since $|x| = 2|\cos \theta| \le 2,$ only the integers -2, -1, 0, +1, +2.

Yeah, my problem is also very interesting and simillar.

Alternatively, for any angle $\theta$ which is a rational multiple of $\pi$, we have that $e^{i \theta}$ and $e^{-i \theta}$ are both algebraic integers. Hence, so is their sum $2 \cos \theta.$ If $\cos \theta$ was to be rational, then so would $2 \cos \theta$ and hence $2 \cos \theta \in \mathbb{Z}$. This gives that $\theta$ is a multiple of $\frac{\pi}{3}.$ Then finish as above.