# In the triangle embedded in the square what is the measure of angle, theta?

Apr 26, 2016

$\theta = \arctan \left(\frac{3 A - \sqrt{A}}{3 A + \sqrt{A}}\right)$

#### Explanation:

Let $s$ represent the side length of the square.

From the right side, we see that $s = \sqrt{A} + \overline{E D}$

From the area of the triangle, we have $A = \frac{s \cdot \overline{E D}}{2}$. Solving for $\overline{E D}$ gives us $\overline{E D} = \frac{2 A}{s}$. Substituting this into the above leaves us with $s = \sqrt{A} + \frac{2 A}{s}$.

With our new equation in $s$ and $A$, we can multiply both sides by $s$ and gather the terms on one side to obtain the quadratic

${s}^{2} - \sqrt{A} s - 2 A = 0$

Applying the quadratic formula gives us

$s = \frac{\sqrt{A} \pm 3 A}{2}$

As we know $s > \sqrt{A}$ we can discard $\frac{\sqrt{A} - 3 A}{2}$, leaving us with

$s = \frac{\sqrt{A} + 3 A}{2}$

We can substitute this value for $s$ into the equation obtained from the right side of the square to obtain

$\frac{\sqrt{A} + 3 A}{2} = \sqrt{A} + \overline{E D}$

$\implies \overline{E D} = \frac{3 A - \sqrt{A}}{2}$

Now, as $\triangle A E D$ is a right triangle, we have

$\tan \left(\theta\right) = \frac{\overline{E D}}{\overline{A D}}$

$= \frac{\overline{E D}}{s}$

$= \frac{3 A - \sqrt{A}}{2} \div \frac{\sqrt{A} + 3 A}{2}$

$= \frac{3 A - \sqrt{A}}{3 A + \sqrt{A}}$

Thus, taking the inverse tangent function of both sides, we get the result

$\theta = \arctan \left(\frac{3 A - \sqrt{A}}{3 A + \sqrt{A}}\right)$