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# Triangle In Circle, Prove An Angle watch

1. (Original post by I hate maths)
I'll challenge that claim with my own improved method.

Draw the diagram to scale and use a protractor.
Wouldn't that be impossible since pi is irrational?
2. (Original post by Y11_Maths)
Aaah I’m so confused. There are so many ways of going about this problem and I don’t even know which is right
I'd recommend following the bear's method because it's the most GCSE and ask him if you get stuck.
3. (Original post by the bear)
angle COB = 2x

the area of triangle ABC = r2/2

the area of triangle OBC = 1/2 of the area of triangle ABC

====> r2/4 = 1/2*OB*OC*Sin2x

etc.
(Original post by the bear)
angle COB = 2x

the area of triangle ABC = r2/2

the area of triangle OBC = 1/2 of the area of triangle ABC

====> r2/4 = 1/2*OB*OC*Sin2x

etc.
Is this right?
I don’t know what to do
4. (Original post by Y11_Maths)

Is this right?
I don’t know what to do
Your last line is wrong. You have r^2 on both sides so they should cancel.
5. (Original post by Notnek)
Your last line is wrong. You have r^2 on both sides so they should cancel.
Oh yes, silly mistake, this time of night haha.
So 1=2sinx.
Where do I go from here? I don’t see where it’s taking me...
6. Deleted
7. (Original post by Y11_Maths)
Oh yes, silly mistake, this time of night haha.
So 1=2sinx.
Where do I go from here? I don’t see where it’s taking me...
You have 1=2sin(2x) - I assume you meant this?

Move the 2 to the other side then you should know what 2x is equal to.
8. (Original post by SumOfSquares)
They did cancel. I think it says 1 on the LHS
No I shouldn’t be left with an r^2 term on the RHS
9. (Original post by Y11_Maths)
No I shouldn’t be left with an r^2 term on the RHS
Yh. I just realized after posting. Sorry.
10. (Original post by SumOfSquares)
Wouldn't that be impossible since pi is irrational?
I was messing around . Standards for proof in school maths are lax but not THAT lax. I think there are other issues if you were to approach the problem that way... You'd have to take a long good look at yourself in the mirror!
11. (Original post by Notnek)
You have 1=2sin(2x) - I assume you meant this?

Move the 2 to the other side then you should know what 2x is equal to.
Yes I did my bad. So sin(2x)= 1/2. Which is 30 degrees. So the other angle is 150 degrees. Since the triangle is isosceles 2x is 30 so x is 15 degrees! Please tell me that I’ve finished????
12. (Original post by Y11_Maths)
Yes I did my bad. So sin(2x)= 1/2. Which is 30 degrees. So the other angle is 150 degrees. Since the triangle is isosceles 2x is 30 so x is 15 degrees! Please tell me that I’ve finished????
Looks like you've finished. Although once you know 2x = 30, you have x = 15 and you're done - I don't know why you mentioned isosceles triangles.

For GCSE you need to know that there are two solutions to sin(x) = a where x<180 and the second can be found by subtracting the first from 180. So another solution is 2x = 150 so x = 75. But your teacher probably doesn't expect you to mention this.
13. (Original post by Notnek)
Looks like you've finished. Although once you know 2x = 30, you have x = 15 and you're done - I don't know why you mentioned isosceles triangles.

For GCSE you need to know that there are two acute solutions to sin(x) = a and the second can be found by subtracting the first from 180. So another solution is 2x = 150 so x = 75. But your teacher probably doesn't expect you to mention this.
Well if triangle OCA has 2 lengths of r then it is isosceles and has 2 angles both of x. Which is why I said this. But I’m tired so I missed your quicker way but it’s fine.

Oh ok thanks for bringing this up! And thank you for helping. Also huge thanks to psc---maths for providing the diagram, this was extremely useful! And thanks to the bear who made it really simple to follow and get the answer. Wohoo!
14. My solution:

From the given statement, area of triangle = r^2/2.

Connecting radii helps to see area of triangle = 1/2 r^2(sina) + 1/2 r^2 sin(180-x) --> [ABC] = r^2sin(a). This means a=30 degrees. <X = 0.5(180-150) = 15 degrees (as the triangle is evidently isosceles).
15. (Original post by thekidwhogames)
My solution:

From the given statement, area of triangle = r^2/2.

Connecting radii helps to see area of triangle = 1/2 r^2(sina) + 1/2 r^2 sin(180-x) --> [ABC] = r^2sin(a). This means a=30 degrees. <X = 0.5(180-150) = 15 degrees (as the triangle is evidently isosceles).
Edit: had to use sin(x) = sin(180-x) but this is in the new GCSE :P
16. (Original post by Y11_Maths)
And thanks to the bear who made it really simple to follow and get the answer. Wohoo!
Woohoo

17. (Original post by the bear)
Woohoo

Are there any limits to the bear’s helpfulness?

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