# Where is the dot?

A dot starts at the point (1,0) and turns anticlockwise. Can you estimate the height of the dot after it has turned through 45 degrees? Can you calculate its height?

*Where is the Dot? printable sheet*

Watch the film below.

Imagine the dot starts at the point $(1,0)$ and turns anticlockwise.

Estimate the height of the dot above the horizontal axis after it has turned through $45^\circ$.

Estimate the angle that the dot needs to turn in order to be exactly $0.5$ units above the horizontal axis.

Show how you can use Pythagoras' Theorem to calculate the height of the dot above the horizontal axis after it has turned through $45^\circ$.

Again, without resorting to Trigonometry, calculate the height of the dot above the horizontal axis after it has turned through $30^\circ$ and $60^\circ$?

Are there any other angles for which you can calculate the height of the dot above the horizontal axis?

The radius of the circle is $1$ unit long.

You can use the radius as the hypotenuse of the right-angled triangles.

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 $45^{\circ}$.

Image

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.

Image

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:

Image

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.

This problem offers students an opportunity to apply Pythagoras' Theorem.

It can also be used as a starting point for trigonometry:

- what happens to the height of the dot during the first $90^{\circ}$ of turn?

- what happens to the height of the dot when it turns beyond $90^{\circ}$?

- what can you say about the horizontal displacement of the dot as it turns through a full circle?