Year 12s: what will be the first way you research your future options? >>

Triangle In Circle, Prove An Angle

So my maths teacher gave me this question at the end of school today and I told him that I would get it done for tomorrow. He is currently "one over on me" so I really want to get the answer back to him so we are even and I have the last laugh. However, the question he gave me has seriously perplexed me and I do not know what to do.

I worked out that bh=r^2 so the area of the triangle is r^2/2 and the area of the circle is pi r^2. But this has taken me nowhere...

Any suggestions?
Thanks in advance

Scroll to see replies

Reply 1

Notnek

Hmm the challenge is to do this without using A Level methods...

Reply 2

Y2_UniMaths

Original post by Notnek

Hmm the challenge is to do this without using A Level methods...

So is this a GCSE question?

Reply 3

Radioactivedecay

Original post by Y11_Maths

So is this a GCSE question?

This is definitely doable for GCSE students, but it requires some thought.

Reply 4

Y2_UniMaths

Original post by Radioactivedecay

This is definitely doable for GCSE students, but it requires some thought.

Help, please?

I'm attempting it...

Original post by Y11_Maths

Help, please?

Try working out CA by splitting the triangle and somehow forming a right angled triangle, then use 1/2ABsinC for the area, then you should see it from there.

Reply 7

username2912782

871AC4A5-6014-4912-8465-6FA13C6A4BB6.jpg.jpeg

Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!

(edited 6 years ago)

Reply 8

username3555092

Original post by Notnek

Hmm the challenge is to do this without using A Level methods...

Which bit in your work did you need A Level maths? If you did something similar to me, you don't actually need to simplify the trig with A Level identities but you could just sub the answer given in to confirm it's the solution.

My gripe with the question is that it doesn't specify angle BAC is less than CBA, so BAC could be 75 instead.

Reply 9

Radioactivedecay

Original post by psc---maths

Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!

I don't think that diagram is correct.

Reply 10

Notnek

Original post by I hate maths

Yes you can just sub it in but I was looking for a nicer GCSE solution that gets the answer directly. You wouldn't see a question like this in a GCSE paper.

Reply 11

Y2_UniMaths

Original post by psc---maths

Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!

The height isn’t r though?

Reply 12

username2912782

Original post by Y11_Maths

The height isn’t r though?

It isn't meant to be the height. It is the line drawn from the centre to B, so it is a radius of length r

Reply 13

username2912782

Original post by psc---maths

It isn't meant to be the height. It is the line drawn from the centre to B, so it is a radius of length r

I'm suggesting you use 1/2absinc for both of the smaller triangles

Reply 14

Radioactivedecay

Original post by Y11_Maths

The height isn’t r though?

Disregard my earlier comment. An easier way would be find the other two sides of the triangle in terms of r and x using SOCAHTOA, and then finding the area of the triangle from there.

Reply 15

username3555092

Original post by Radioactivedecay

I don't think that diagram is correct.

I think it is correct but the original triangle is easy enough to work with without splitting.

Reply 16

username2912782

Original post by I hate maths

I think it is correct but the original triangle is easy enough to work with without splitting.

Wrt the other way would that not require 2sinxcosx=sin2x. I was giving a gcse level answer

Reply 17

Y2_UniMaths

Original post by psc---maths

Find the area of the two smaller triangles in terms of x using 1/2absinc then find triangle/circle and equate this to 1/2pi. Reminder that sin(180-2x)=sin(2x): think about the graph of sin!

The diagram doesn’t look accurate? Where did you get 180-2x from? The top vertice of the triangle is not vertically above the centre and how can OC be r? I’m really confused

Reply 18

Y2_UniMaths

Original post by Radioactivedecay

Disregard my earlier comment. An easier way would be find the other two sides of the triangle in terms of r and x using SOCAHTOA, and then finding the area of the triangle from there.

Isn’t the area of the triangle r^2/2, or am I wrong?

Reply 19

username3555092

Original post by psc---maths

Wrt the other way would that not require 2sinxcosx=sin2x. I was giving a gcse level answer

As I was saying to Notnek, it wouldn't require this as the question gives you the value of x, so you can sub it in and confirm it's the solution without knowledge of identities. People tend to have an aversion for this kind of solution but I think it's a useful way to do some problems.