This is about compass and straightedge constructions, although I suspect nothing changes if we expand to real numbers algebraic over the rationals.

Anyway, let $E$ be the "constructible numbers," meaning the smallest subfield of the reals such that, if $x > 0, x \in E,$ then $\sqrt x \in E.$ So the elements are towers of square roots, add and mix those together.

Next, suppose we have real numbers $\alpha, \beta$ such that $\sin \alpha \in E,\sin \beta \in E. $ These are constructible angles written in radians.

It is easy to show with Hermite-Lindemann that, for $m,n \in \mathbb Z,$ that $$ m \alpha + n \beta \notin E $$ unless $$ m \alpha + n \beta = 0.$$ Divide though by another integer, we get the same statement for rational coefficients. Divide by the first, we have the statement for $\alpha + r \beta$ with $r \in \mathbb Q.$

Question: if $x \in E$ but $x \notin \mathbb Q,$ also $\sin \alpha \in E,\sin \beta \in E $ and $\alpha, \beta \neq 0,$ can we conclude that $$ \color{magenta}{ \alpha + x \beta \notin E ?} $$

Caution: it is all nice if $\alpha, \beta$ are rational or algebraic multiples of $\pi.$ However, perfectly good angles such as $\arctan 2$ are involved, this being a transcendental multiple of $\pi.$