Age 7 to 11
Pupils from Stradbroke Primary School sent
in some well-thought out solutions to this problem. Roni explains
how they worked out how much further (from $20^\circ$) the point
would have to turn to finish the same height above or below the
1. A whole circle is $360^\circ$, half a circle is
$180^\circ$.$180^\circ$ take away $20^\circ$ and then take away
another $20^\circ$ gives the answer of $140^\circ$.
2. The first $20^\circ$ is above the x line and the third
$20^\circ$ is under the x line so it's always $180^\circ$.
3. A whole circle is $360^\circ$, you take away $20^\circ$ and then
another $20^\circ$. 20 add 20 equals 40. 360 take away 40 equals
So, Roni has concluded that there are three
other points where the dot would be the same distance from the
horizontal axis as it is for $20^\circ$. Another way of looking at
it would be to say that the dot itself would have to be at
$20^\circ$, $160^\circ$, $200^\circ$and $340^\circ$.
George tells us how we can work out how far
a point at any position on the circle would have to turn to finish
the same height above or below the horizontal axis. He agrees that
there will be three other solutions:
1. Double the first degree and take it away from $180^\circ$.
2. Add $180^\circ$ to the your starting degree.
3. Double the starting degree and take it away from