Come on then people, let's see who can get it:
(http://home.bt.com/images/maths-exam-question-136401423147603901-151103122131.jpg)
A question that appeared in a further maths exam has prompted debate over whether it was too hard to answer.
The multiple-choice poser was given to 17- and 18-year-old Year 12 students in Australia and asked them to find the degree of angle between two 50 cent coins with 12 sides each.
Students had a choice of answers: 12, 30, 36, 60 or 72 degrees. Many took to social media to complain the question was simply too difficult.
"That exam wasn't there to test us, it was to trick, no one could have prepared for that," one student wrote.
Another said: "The 50 cent coin question makes no sense to me as it asked us to find the angle, but it gave us no other angles or lengths to enter in the formulas so there was simply no way about it."
However, others claimed the question was too easy for a further maths exam.
A teacher, Darren Adams wrote: "And this stumped VCE students? Gimme a break! My Year 5s could have worked it out."
Pothiraju Seetharam added: "It is easy. Overthinking did them in."
I'll post up the answer once we've had a few replies in, but I was surprised the students struggled with this as much as they seem to have done.
:-\
I'm probably about to show why the Maths Department at school suggested I do French at Higher rather than Maths but ...
If each coin is effectively 12 identical isosceles triangles then the angle at the centre point of the coin is a twelfth of 360o or 30o.
Therefore the angles at the bottom are 180o minus 30o divided by two or 75o.
Therefore the total of the angles "inside" the coin at the point of juncture are 4 times 75o or 300o.
360o less 300o = 60o
Theta = 60o
Quod Erat Demonstrandum ... or does that prove my Latin is as bad as my Maths?
I got 60 too...
I refuse to believe this is a real story. The implications on the maths abilities of today's youth are too worrying to consider if it is true.
The coin has 12 sides. Ergo the degree of change between each side is
Degrees in a circle / Number of sides
360 / 12
30.
This is the outside angle. (The interior angle must be 150)
Two coins are aligned along one side. Therefore the line of which the 30 is taken must be reflected.
2 x 30 =60.
Quote from: mad lemmey on 03 November 2015, 11:54:53 PM
I got 60 too...
I got 60 without doing any of that fancy brainwork.
Yup 60, but i calculated a bit differently.
Difference between the side standing flat and the side where the coins touch is 90°, and has 3 "waypoints" (wich are at equal distances because the sides are of equal length)
So 90°/3 = 30
2 x 30° = 60°
Quote from: Last Hussar on 04 November 2015, 12:06:00 AM
I refuse to believe this is a real story. The implications on the maths abilities of today's youth are too worrying to consider if it is true.
Everyone from Huffpo to the Daily Wail are reporting it as true ... whether you believe what you read in the papers? .... well ... :)
I got 60.
Fairly straight forward. Coin has 12 sides. Each side must turn 30 degrees. the two sides are turning away from each other so 2x 30 = 60
Seems like no one had a much of a problem with that then!
Quote from: petercooman on 04 November 2015, 12:10:16 AM
Difference between the side standing flat and the side where the coins touch is 90°, and has 3 "waypoints" (wich are at equal distances because the sides are of equal length)
So 90°/3 = 30
2 x 30° = 60°
That's the way I did it as well.
Here's the official answer:
How to work out the answer
The answer to the question is 60 degrees. There are a few ways to come to the correct solution, but here's how we worked it out:
1. Calculate the sum of the interior angles of the coin. The formula for this is (Number of sides x 180) – 360, so the sum of the interior angles of the 12-sided coin is (12x180)-360=1800.
2. The coin is a regular 12-sided polygon, so all its internal angles are equal. To find the value of one internal angle, divide the sum of interior angles by the number of sides, so 1800/12=150.
3. A full circle is divided into 360 degrees. In the question the angle θ comprises part of a circle along with one internal angle from each of the coins. As one internal angle is 150 degrees, the two adjacent angles add up to 300 degrees, leaving θ with a value of 360-300=60 degrees.
Rob Muldoon, a 'famous' NZ Prime Minister, once said "Any New Zealander choosing to emigrate to Australia is raising the average IQ of both countries" ;)
Quote from: Leon on 03 November 2015, 11:16:31 PM
Another said: "The 50 cent coin question makes no sense to me as it asked us to find the angle, but it gave us no other angles or lengths to enter in the formulas so there was simply no way about it."
I think this comment from a student highlights the key to the problem, they had prepared to use formulas, not to think ;)
Quote from: paulr on 04 November 2015, 02:39:46 AM
Rob Muldoon, a 'famous' NZ Prime Minister, once said "Any New Zealander choosing to emigrate to Australia is raising the average IQ of both countries" ;)
I now have to like this sort of comment :D
Cheers
Ian
I just saw an equilateral triangle, and just thought it has to be 60 degrees.
Cheers - Phil
You can look at it and see that it is 60 degrees! Even I with my grade C at maths O level worked it out in less than a minute.
Quote from: Leon on 04 November 2015, 01:14:17 AM
Seems like no one had a much of a problem with that then!
That's the way I did it as well.
Here's the official answer:
How to work out the answer
The answer to the question is 60 degrees. There are a few ways to come to the correct solution, but here's how we worked it out:
1. Calculate the sum of the interior angles of the coin. The formula for this is (Number of sides x 180) – 360, so the sum of the interior angles of the 12-sided coin is (12x180)-360=1800.
2. The coin is a regular 12-sided polygon, so all its internal angles are equal. To find the value of one internal angle, divide the sum of interior angles by the number of sides, so 1800/12=150.
3. A full circle is divided into 360 degrees. In the question the angle θ comprises part of a circle along with one internal angle from each of the coins. As one internal angle is 150 degrees, the two adjacent angles add up to 300 degrees, leaving θ with a value of 360-300=60 degrees.
Well, that's too much work for something so simple ;D ;D ;D ;D
Logic speeds up everything!
To be honest I judged by eye that it was 60o then worked out how you proved that.
Given that, as I said, when it came to sitting Highers the Maths Department suggested I would be better doing French (to be fair the French Department suggested I did Maths but they lost! :) ) it doesn't say much for the teaching of Maths in Australia that this was a problem.
Alas, this seems to me to be one more downside of teaching children to pass exams rather than teaching them the subject.
For my History Higher course the syllabus covered Europe from 1816-1913 and Scotland from the 1600s to the 1800s. Having studied previous papers I prepared four topics in depth, needing to answer three questions. None of them came up!! I was able to get an "A" by giving answers on the boundaries of the Roman Empire, the armies of Gustavus Adolphus and the Code Napoleon because I was taught History not Exams!
</rant> :)
Easiest way was flat surface 180 degree two 50 cent coins meeting at the 90 degree so three angles = 60 degree each.
Simples
Sean