Find the angle between two vectors in a trapezoid made up of 5 equilateral triangles I am working on a problem where I have to determine the angle between vector AC and AD. The trapezoid is made up of 5 equilateral triangles.
I know equilateral triangles have equal sides and an angle of 60 degrees. But I am not sure how to apply that knowledge to solve this problem.

 A: Perhaps there is a more straightforward way, but here is my approach. I’ll leave you the details of actually solving the problem.
Use the law of cosines to calculate the length of $AC$. In this case, it would be:
$|AC|^2=|AD|^2+|CD|^2-2|AD||CD|\cos\frac{\pi}{3}$
The lengths may be determined by using the fact that the triangles are equilateral.
Then the law of sins may be used ($ac$ is the angle across from $AC$):
$\frac{\sin ac}{|AC|}=\frac{\sin cd}{|CD|}$
The angle you are solving for is $cd$.
This problem may also be solved using the dot product property $A\cdot B=||A||||B||\cos\theta$.
A: The given figure reminds me the analytic geometry lesson. Let's give coordinates to points. Let $A=(0,0)$ and $D=(3,0)$ so that $\overleftrightarrow{AB}$ is $x$-axis. $B$ lies on the unit circle. Since $\angle BAD=60^{\circ}$, $B=(\cos 60^{\circ},\sin 60^{\circ})=(\frac12,\frac{\sqrt3}{2})$. Hence, $C=(2+\frac12,\frac{\sqrt3}{2}+0)=(\frac52,\frac{\sqrt3}{2}).$ Writing in terms of vectors, $\vec{AC}=\langle \frac52,\frac{\sqrt3}{2} \rangle$ and $\vec{AD}=\langle 3,0\rangle$ since $A$ is the origin. Finally, letting $\theta=\angle CAB$, we have
$$\theta=\arccos\left(\frac{\vec{AC}\cdot\vec{AD}}{\Vert\vec{AC}\Vert\Vert\vec{AD}\Vert}\right)=\arccos\left(\frac{\frac52.3+0}{\sqrt{(\frac52)^2+(\frac{\sqrt3}{2})^2}\sqrt{3^2+0^2}}\right)=\arccos\left(\frac{5}{2\sqrt7}\right)\approx 19.11^{\circ}$$
A: You can use elementary trigonometry. Drop a perpendicular from $C$ onto $AD$ so the foot of the perpendicular is $X$.
If the side of each equilateral triangle is $2$, say, the height $CX$ is $\sqrt{3}$ and the base $AX$ is $5$. Therefore the required angle is $$\tan^{-1}\left(\frac{\sqrt{3}}{5}\right)=19.11^o$$
