I was asked to solve for the $\theta$ shown in the figure below.
My work:
The $\Delta FAB$ is an equilateral triangle, having interior angles of $60^o.$ I don't think $\Delta HIG$ and $\Delta DEC$ are right triangles.
So far, that's all I know. I'm confused on how to get $\theta.$ How do you get the $\theta$ above?
Hints:
Triangle HIG, and triangle DCE are isosceles triangles, with $\angle HIG = \angle DCE = 90^\circ - 60^\circ = 30^\circ$.
In isosceles triangles, the base angles are congruent, so $\angle IHG = \angle HGI = \dfrac{180^\circ - 30^\circ}{2} = 75^\circ.$
Similarly, for $\angle CED$ and $\angle CDE$, they too are $75^\circ$.
This answer makes use of analytic geometry, as an alternative to other answers.
Setting a cartesian coordinate system with origin in $A$ and $x$-axis parallel to $AB$ and $y$-axis parallel to $AH$ you have:
$y_F=\frac{\sqrt{3}}{2}X$
$\tan\frac{\theta}{2}=\frac{X/2}{y_H-y_F}=\frac{1}{2}\frac{1}{1-\frac{\sqrt{3}}{2}}=\frac{1}{2-\sqrt{3}}$
$\theta=2\arctan\frac{1}{2-\sqrt{3}}=150^\circ$
First, some general advice
With that said, an argument follows, assuming that the large quadrilateral is a square.
Indeed, $\triangle HIG$ and $\triangle DEC$ are not right triangles.
Observe that $\angle I$ and $\angle A$ are complementary (i.e. they add up to a right angle). But $\angle A$ measures $60^{\circ}$, and so $\angle I$ must measure $30^{\circ}$. By similar reasoning, $\angle C$ also measures $30^{\circ}$.
By the side-angle-side congruence relation (SAS), you know that $\triangle GIH \cong \triangle ECD$. We just argued that the middle angles are $30^{\circ}$, and from the diagram, we have that $GI = IH = EC = CD = x$. Indeed, not only are the two triangle congruent, they are both isosceles! This means that $$ m\angle G = m\angle H \qquad\text{and}\qquad m\angle G + m\angle H + m\angle I = 180^{\circ},$$ since the angle sum of a triangle is $180^{\circ}$. Combining these, and solving for either $m\angle G$ or $m\angle H$, we obtain $$ m\angle G = m\angle H = 75^{\circ}.$$ By similar reasoning $$ m\angle E = m\angle D = 75^{\circ}.$$ The two unlabeled angles are equal, as they are each the complement of an angle that is $75^{\circ}$. Thus the two unlabeled angles each measure $15^{\circ}$. Finally, in the unlabeled triangle, we know that the angle sum must, again, be $180^{\circ}$. Since we know that two of the angles are $15^{\circ}$ each, we solve $$ \theta + 15^{\circ} + 15^{\circ} = 180^{\circ}$$ in order to obtain $\theta = 150^{\circ}$.
If the figure is not a square, then things are slightly more complicated (though not really):
Since the quadrilateral $ABDC$ is equilateral (i.e. all four sides are of the same length), it must be a rhombus, so opposite sides are parallel. This implies that $$ \alpha + \beta + 120^{\circ} = 180^{\circ} $$ (make sure you understand why). From this, it follows that $$ \frac{1}{2}(\alpha + \beta) = 30^{\circ}.$$ Note that $\triangle EAC$ is isosceles (it has two sides that are congruent), from which it follows that the two remaining unknown angles have measure $$ 90^{\circ} - \frac{\alpha}{2}. $$ Similarly, the two unlabeled angles in $\triangle EBD$ have measure $$ 90^{\circ} - \frac{\beta}{2}. $$ Again using the fact that $\overline{AC} \parallel \overline{BD}$, it follows that $$ \left( 90^{\circ} - \frac{\alpha}{2} \right) + \left( 90^{\circ} - \frac{\beta}{2} \right) + \gamma + \delta = 180^{\circ}. $$ Because this turns out to be useful, we can rearrange this to get $$ \gamma + \delta = \frac{1}{2} (\alpha + \beta) = 30^{\circ}$$ But the angle sum of a triangle is $180^{\circ}$, and so $$ \theta = 180^{\circ} - (\gamma + \delta) = 180^{\circ} - 30^{\circ} = 150^{\circ}.$$ Note that this is the same answer we got above, which goes to show that it is often possible to get the right answer for reasons that are not quite right.
