Angles involving trigonometric Ratios Worded Problem

[fresh_42]

I don't think they are wrong. I have $x:=AB$ and thus $AC=3x$ opposite of $\beta = \angle (ABC) = 60°-\theta$ and $x=AB$ opposite of $\gamma = \angle (BCA) =\theta$.

You can get rid of $x$ because you're not interested in its value. $\sin \theta$ is what we're looking for.