UnTangle without calculator

Notice that $\tan \theta$ is an increasing function when $0^\circ < \theta < 90^\circ$ .

Note that $\tan(75^\circ) = \tan(30^\circ + 45^\circ)$ . We apply the compound angle formula , $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ to get $\tan(75^\circ) = 2 + \sqrt3$ .

So we just need to determine whether $\dfrac{11}3$ less than or greater than $2 + \sqrt3$ .

In other words,
If $\tan \theta = \dfrac{11}{3}$ is less than $2 + \sqrt3$ , then $0^\circ < \theta < 75^\circ$ .
If $\tan \theta = \dfrac{11}{3}$ is greater than $2 + \sqrt3$ , then $75^\circ < \theta < 90^\circ$ .

$\begin{aligned} 11/3 & \quad ? \quad & 2 + \sqrt3 \\ 5/3 & \quad ? \quad & \sqrt3\\ 5 & \quad ? \quad & 3 \sqrt3 \\ 25& \quad ? \quad & 27 \end{aligned}$

So "?" should be " $<$ ". Thus, $\boxed{\theta < 75^\circ}$ .

We will calculate $\tan 2\theta$ and compare it to $\tan 150^\circ = -\tan 30^\circ = -\tfrac13\sqrt 3$ .

$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2\cdot 11/3}{1 - (11/3)^2} = \frac{66}{9 - 121} = -\frac{66}{112} = -\frac{33}{56}.$

Now $-33/56 < -\tfrac13\sqrt 3$ if and only if $(33/56)^2 > 1/3$ . We check: $\left(\frac{33}{56}\right)^2 = \frac{1089}{3136} > \frac{1089}{3267} = \frac13.$

We conclude that indeed $\tan 2\theta = -33/56 < -\tfrac13\sqrt 3 = \tan 150^\circ$ . It follows that $\boxed{\tan\theta < 75^\circ}$ .