Solving Triangles - Part 8

Putting $BC=2$ as a base, we note that the point $A$ is along the side $A"B$ and $D$ is along the side $A"C$ and an isosceles $\triangle A"BC$ . As we move the line $AD$ downward, both $AB$ and $DC$ shorten until $DC=0$ , then the point $A$ is $A'$ . Similarly, as we move the line $AD$ upward, both $AB$ and $DC$ lengthen until $DC=AB$ , then the point $A$ is $A"$ . Therefore the range of $AB$ is given by:

$\begin{aligned} A'B < & AB < A"B \\ 2(2\cos 75^\circ) < & AB < \frac {\frac 22}{\cos 75^\circ} & \small \color{#3D99F6} \text{Note that }\cos 75^\circ = \cos (30^\circ + 45^\circ) = \frac {\sqrt 6 - \sqrt 2}4 \\ 4 \left(\frac {\sqrt 6-\sqrt 2}4\right) < & < AB < \frac 4{\sqrt 6 - \sqrt 2} \\ \sqrt 6-\sqrt 2 < & < AB < \sqrt 6 + \sqrt 2 \end{aligned}$

Therefore, $AB \in \boxed{\left(\sqrt 6 - \sqrt 2, \sqrt 6 + \sqrt 2\right)}$ .

With the given information, quadrilateral $ABCD$ can look like the one pictured on the left:

but $AB$ and $CD$ can shrink or grow and still fit the given restrictions. As $CD$ approaches $0$ (middle picture), we have a triangle which by law of sines $AB = \frac{2 \sin 30°}{\sin 75°} = \sqrt{6} - \sqrt{2}$ , and as $AD$ approaches $0$ (right picture), we have a triangle which by law of sines $AB = \frac{2 \sin 75°}{\sin 30°} = \sqrt{6} + \sqrt{2}$ .

The triangles above are not included as quadrilaterals, therefore the range of $AB$ is $\boxed{(\sqrt{6} - \sqrt{2}, \sqrt{6} + \sqrt{2})}$ .