As is customary, I spotted a math problem and had to solve it. It’s like a sickness. Anyway, here’s the answer to that math problem that stumped Year 12 math students in Australia!
The problem goes like this:
There are 2 things you need to know to solve this problem:
- Angles inside of a triangle add up to 180 degrees.
- Angles in a circle add up to 360 degrees.
Because the center of the 50 cent coin encompasses 360 degrees, with edges that are broken off into 12 segments, you can divide 360 by 12 to get the angle of each triangle inside the coin. Let’s call that angle “alpha.”
360 / 12 = 30. Alpha is 30 degrees.
Now, you’ve made 12 triangles when you divided the coin into 12 segments. Each triangle has all three of its angles adding up to 180 degrees. So you can solve for the other 2 angles in the triangle, which will be equal. You know they’re equal because all of the triangles are the same. We’ll call the bottom two angles “beta.”
With a little bit of algebra, you can easily figure out what beta is. 180 = 2B + 1A, because you have 2 beta’s and 1 alpha in each triangle. You know what alpha is. Now you only have one variable. Time to plug and chug!
180 = 2B + 30
150 = 2B
75 = B
Beta is 75 degrees!
Now notice something about X. X is the angle between those two coins, but it’s also part of a circle.
Circles have 360 degrees in them. And we already know how many degrees are in all of the other divisions of this circle, because we found beta!
Notice that the circle is made of 4 beta’s and one X. Time for algebra again!
So 360 = X + 4(B)
360 = X + 4(75)
360 – 4(75) = X
360 – 300 = X
60 = X
And that’s the answer!
I hope this was helpful. Comment below!
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