Maths question from Cambridge natural sciences past paper

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knoxode
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I really don't know where to start with this question to be honest.
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ghostwalker
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(Original post by knoxode)
I really don't know where to start with this question to be honest.
Starting point is going to depend on what you know. So, first off, what do you know about the angles in a regular polygon?
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knoxode
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(Original post by ghostwalker)
Starting point is going to depend on what you know. So, first off, what do you know about the angles in a regular polygon?
The interior angle of a regular polygon is equal to the sum of the interior angles, divided by the number of sides.
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ghostwalker
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(Original post by knoxode)
The interior angle of a regular polygon is equal to the sum of the interior angles, divided by the number of sides.
OK. Can't usefully use that.

So, most direct root, IMO, is to look at the exterior angle.

If we follow the line of the polygon round, it goes through a full 360 degrees, and this is n lots of the exterior angle, one after the other.

Hence the exterior angle equals 360/n, and that will be an angle in the isosceles triangle. Can you take it from there?
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knoxode
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(Original post by ghostwalker)
OK. Can't usefully use that.

So, most direct root, IMO, is to look at the exterior angle.

If we follow the line of the polygon round, it goes through a full 360 degrees, and this is n lots of the exterior angle, one after the other.

Hence the exterior angle equals 360/n, and that will be an angle in the isosceles triangle. Can you take it from there?
So perhaps the other to angles in the isosceles triangles (that are not x) will be equal to n/360 - 180?
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ghostwalker
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(Original post by knoxode)
So perhaps the other to angles in the isosceles triangles (that are not x) will be equal to n/360 - 180?
Not sure how you get that.

Each of the not-x angles is 360/n, since one is an exterior angle and it's isosceles.

Hence for the whole triangle:

x + (360/n) +(360/n) = 180

And rearrange.
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