If tan2θ = 4/3 and 2θ is acute, then tanθ equals:

Put $t$ = tanθ in tan2θ = $2t/(1-t^2):3(2t)=4(1-t^2)$, so $4t^2+6t-4=0$, i.e. $2t^2+3t-2=(2t-1)(t+2)=0$. The roots are $t=1/2$ and $t=-2$, and since 2θ is acute so is θ, forcing a positive tangent: $t=1/2$ (θ ≈ 26.57°, 2θ ≈ 53.13°). The rejected root supplies −2, its size supplies 2, and simply halving 4/3 gives 2/3.