Question

find the tangent of $\angle e$.

triangle def with right angle at f, ef = $6\sqrt{33}$, fd = $6\sqrt{33}$, ed = $6\sqrt{66}$

write your answer in simplified, rationalized form. do not round.
$\tan(e) = \square$

Explanation:

Step1: Recall tangent definition

In a right triangle, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ for an acute angle $\theta$. For $\angle E$ in right triangle $EFD$ (right-angled at $F$), the side opposite $\angle E$ is $FD$, and the side adjacent to $\angle E$ is $EF$.

Step2: Identify side lengths

From the diagram, $FD = 6\sqrt{33}$ and $EF = 6\sqrt{33}$.

Step3: Calculate $\tan(E)$

Using the tangent ratio: $\tan(E) = \frac{FD}{EF} = \frac{6\sqrt{33}}{6\sqrt{33}} = 1$? Wait, no, wait. Wait, maybe I mixed up. Wait, let's check again. Wait, in triangle $EFD$, right-angled at $F$. So angle at $F$ is right angle. So angle $E$: the opposite side to $E$ is $FD$, adjacent is $EF$. Wait, but $FD$ is $6\sqrt{33}$, $EF$ is $6\sqrt{33}$? Wait, no, wait the hypotenuse is $ED = 6\sqrt{66}$. Wait, let's verify with Pythagoras: $(6\sqrt{33})^2 + (6\sqrt{33})^2 = 36*33 + 36*33 = 72*33 = 2376$. And $(6\sqrt{66})^2 = 36*66 = 2376$. So yes, it's an isoceles right triangle? Wait, but then $\tan(E)$ would be opposite over adjacent, which is $FD/EF = (6\sqrt{33})/(6\sqrt{33}) = 1$? But that seems too simple. Wait, maybe I got the sides wrong. Wait, no, angle $E$: in triangle $EFD$, right-angled at $F$. So angle $E$: the sides: $EF$ is horizontal, $FD$ is vertical, $ED$ is hypotenuse. So for angle $E$, the opposite side is $FD$ (vertical), adjacent is $EF$ (horizontal). So $FD = 6\sqrt{33}$, $EF = 6\sqrt{33}$. So $\tan(E) = \frac{FD}{EF} = \frac{6\sqrt{33}}{6\sqrt{33}} = 1$. Wait, but let's check again. Wait, maybe the labels are different. Wait, the triangle is labeled $E$, $F$, $D$ with right angle at $F$. So $E$ to $F$ is $6\sqrt{33}$, $F$ to $D$ is $6\sqrt{33}$, $E$ to $D$ is $6\sqrt{66}$. So yes, it's a right isoceles triangle, so angle $E$ is 45 degrees, tangent of 45 is 1. So that's correct.

Wait, but maybe I made a mistake. Wait, let's re-express:

$\tan(E) = \frac{\text{opposite}}{\text{adjacent}} = \frac{FD}{EF} = \frac{6\sqrt{33}}{6\sqrt{33}} = 1$. Yes, that's correct.

Answer:

$\boxed{1}$