# A composite geometric shape of triangle and rectangle is given with the proportionate relation given in the figure. Find angle alpha and theta?

## May 2, 2016

$\theta = \arctan \left(\frac{2}{3}\right) \approx {33.69}^{\circ}$

$\alpha = {90}^{\circ} - \arctan \left(\frac{1}{3}\right) - \arctan \left(\frac{1}{4}\right) \approx {57.53}^{\circ}$

#### Explanation:

Labeling the picture as follows (with the additional constructed segment $\overline{A E}$): From the right triangle $\triangle B C F$, given $\overline{C F} = \frac{b}{2} = h$ and $\overline{B F} = \frac{2 h}{3}$, we have $\tan \left(\theta\right) = \frac{\frac{2 h}{3}}{h} = \frac{2}{3}$

Applying the inverse tangent function gives us $\theta$:

$\theta = \arctan \left(\frac{2}{3}\right) \approx {33.69}^{\circ}$

Note that $\overline{E C} = \overline{E F} - \overline{C F} = \overline{A D} - \overline{C F} = \frac{2 b}{3} - \frac{b}{2} = \frac{b}{6} = \frac{h}{3}$

Together with $\overline{A E} = h$, we have the angle $\angle E A C = \arctan \left(\frac{\frac{h}{3}}{h}\right) = \arctan \left(\frac{1}{3}\right)$

Additionally, as $\overline{B D} = \frac{h}{3}$ and $\overline{A D} = \frac{2 b}{3} = \frac{4 h}{3}$, we have the angle $\angle D A B = \arctan \left(\frac{\frac{h}{3}}{\frac{4 h}{3}}\right) = \arctan \left(\frac{1}{4}\right)$

Noting that $\angle E A D$ is a right angle, we can now obtain $\alpha$:

$\alpha = \angle E A D - \angle E A C - \angle D A B$

$= {90}^{\circ} - \arctan \left(\frac{1}{3}\right) - \arctan \left(\frac{1}{4}\right) \approx {57.53}^{\circ}$