Triangle In Circle, Prove An Angle

I'll challenge that claim with my own improved method.

Draw the diagram to scale and use a protractor.

Aaah I’m so confused. There are so many ways of going about this problem and I don’t even know which is right

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.

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.

But if we already

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.

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...

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

Wouldn't that be impossible since pi is irrational?

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????

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.

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).

And thanks to @the bear who made it really simple to follow and get the answer. Wohoo!

Woohoo