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'Where Is the Dot?' printed from https://nrich.maths.org/
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.