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Pythagoras' theorem states that $x^2+x^2=1^2$, i.e. $x^2=\frac{1}{2}$ so $x=\frac{1}{\sqrt{2}}$.
The next part of the question asked for similar ways of calculating the height of the dot after it had turned through $30^{\circ}$ and $60^{\circ}$.
Consider the right angled triangle we obtain after turning through $30^{\circ}$. If we reflect this triangle in the horizontal axis we obtain an equalateral triangle with sides of length $1$ as shown.
This implies that the height of the dot must be $\frac{1}{2}$. We deduce that at $60^{\circ}$ we end up with the following triangle:
By Pythagoras, the height of the dot must satisfy the equation $x^2+(\frac{1}{2})^2=1^2$ which implies that $x=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$.
By symmetry, now that we know the height of the dot for angles of $30^{\circ}$, $45^{\circ}$ and $60^{\circ}$, we can state the height for angles which are a multiple of $30^{\circ}$, $45^{\circ}$ or $60^{\circ}$. See if you can list these angles and the corresponding heights of the dot.
Where Is the Dot?
Age 14 to 16
Challenge Level 
The first part of this problem was answered correctly by a
number of people. As the film suggests, we can view the dot as a
vertex on a right angled triangle whose hypotenuse has length one.
The angle $45^{\circ}$ is special because at this point the
triangle is equalateral. Let $x$ denote the height of the dot after
it has turned through $45^{\circ}$.
Pythagoras' theorem states that $x^2+x^2=1^2$, i.e. $x^2=\frac{1}{2}$ so $x=\frac{1}{\sqrt{2}}$.
The next part of the question asked for similar ways of calculating the height of the dot after it had turned through $30^{\circ}$ and $60^{\circ}$.
Consider the right angled triangle we obtain after turning through $30^{\circ}$. If we reflect this triangle in the horizontal axis we obtain an equalateral triangle with sides of length $1$ as shown.
This implies that the height of the dot must be $\frac{1}{2}$. We deduce that at $60^{\circ}$ we end up with the following triangle:
By Pythagoras, the height of the dot must satisfy the equation $x^2+(\frac{1}{2})^2=1^2$ which implies that $x=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$.
By symmetry, now that we know the height of the dot for angles of $30^{\circ}$, $45^{\circ}$ and $60^{\circ}$, we can state the height for angles which are a multiple of $30^{\circ}$, $45^{\circ}$ or $60^{\circ}$. See if you can list these angles and the corresponding heights of the dot.
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